Mixing
Solving y = 1/4 log(2) (8x – 56)^16 – 12
$begingroup$
Hello I would like to know if whether I simplified y = 1/4 log2 (8x – 56)^16 – 12 correctly or not. Here are my steps:
12 = log2 (8x – 56)^4 => I moved the 1/4 to the power using the log laws. 1/4 x 16 equals 4. I also moved the -12 to the other side which equals +12.
Log2(8x)^4 – log2(56)^4 = 12 => Since (8x-56) has the same log base I rewrote it as two terms.
Log2(8x/56)^4 = 12 => Using the log laws I divided the 8x with 56.
Log2(x/7)^4 = 12 => I simplified the fraction.
Log2(x^4/7^4 ) = 12 => I applied the power of 4 to the fraction.
2^12 = (x^4/7^4 ) => I converted the logarithm to an exponential equation
4096/2401 = x^4 => I carried the 2401 (7^4) to the other side. Dividing the 2^12 with 7^4.
∜(4096/2401) = x => I fourth rooted both sides.
x = 8/7 = I arrived with this fraction.
My question is, did I solve the equation correctly? Thank you!
proof-verification logarithms
asked Jan 30 at 17:56
moemoe
82
$endgroup$
add a comment |
$begingroup$
Hello I would like to know if whether I simplified y = 1/4 log2 (8x – 56)^16 – 12 correctly or not. Here are my steps:
12 = log2 (8x – 56)^4 => I moved the 1/4 to the power using the log laws. 1/4 x 16 equals 4. I also moved the -12 to the other side which equals +12.
Log2(8x)^4 – log2(56)^4 = 12 => Since (8x-56) has the same log base I rewrote it as two terms.
Log2(8x/56)^4 = 12 => Using the log laws I divided the 8x with 56.
Log2(x/7)^4 = 12 => I simplified the fraction.
Log2(x^4/7^4 ) = 12 => I applied the power of 4 to the fraction.
2^12 = (x^4/7^4 ) => I converted the logarithm to an exponential equation
4096/2401 = x^4 => I carried the 2401 (7^4) to the other side. Dividing the 2^12 with 7^4.
∜(4096/2401) = x => I fourth rooted both sides.
x = 8/7 = I arrived with this fraction.
My question is, did I solve the equation correctly? Thank you!
proof-verification logarithms
asked Jan 30 at 17:56
moemoe
82
$endgroup$
$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47
add a comment |
$begingroup$
Hello I would like to know if whether I simplified y = 1/4 log2 (8x – 56)^16 – 12 correctly or not. Here are my steps:
12 = log2 (8x – 56)^4 => I moved the 1/4 to the power using the log laws. 1/4 x 16 equals 4. I also moved the -12 to the other side which equals +12.
Log2(8x)^4 – log2(56)^4 = 12 => Since (8x-56) has the same log base I rewrote it as two terms.
Log2(8x/56)^4 = 12 => Using the log laws I divided the 8x with 56.
Log2(x/7)^4 = 12 => I simplified the fraction.
Log2(x^4/7^4 ) = 12 => I applied the power of 4 to the fraction.
2^12 = (x^4/7^4 ) => I converted the logarithm to an exponential equation
4096/2401 = x^4 => I carried the 2401 (7^4) to the other side. Dividing the 2^12 with 7^4.
∜(4096/2401) = x => I fourth rooted both sides.
x = 8/7 = I arrived with this fraction.
My question is, did I solve the equation correctly? Thank you!
proof-verification logarithms
asked Jan 30 at 17:56
moemoe
82
$endgroup$
Hello I would like to know if whether I simplified y = 1/4 log2 (8x – 56)^16 – 12 correctly or not. Here are my steps:
12 = log2 (8x – 56)^4 => I moved the 1/4 to the power using the log laws. 1/4 x 16 equals 4. I also moved the -12 to the other side which equals +12.
Log2(8x)^4 – log2(56)^4 = 12 => Since (8x-56) has the same log base I rewrote it as two terms.
Log2(8x/56)^4 = 12 => Using the log laws I divided the 8x with 56.
Log2(x/7)^4 = 12 => I simplified the fraction.
Log2(x^4/7^4 ) = 12 => I applied the power of 4 to the fraction.
2^12 = (x^4/7^4 ) => I converted the logarithm to an exponential equation
4096/2401 = x^4 => I carried the 2401 (7^4) to the other side. Dividing the 2^12 with 7^4.
∜(4096/2401) = x => I fourth rooted both sides.
x = 8/7 = I arrived with this fraction.
My question is, did I solve the equation correctly? Thank you!
proof-verification logarithms
proof-verification logarithms
asked Jan 30 at 17:56
moemoe
82
asked Jan 30 at 17:56
moemoe
82
edited Jan 30 at 23:49
moe
asked Jan 30 at 17:56
moemoe
82
asked Jan 30 at 17:56
moemoe
82
asked Jan 30 at 17:56
moemoe
82
82
$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47
add a comment |
$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47
$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You had a good start. $12 = log_2 (8x – 56)^4$ can be rewritten as $12 = 4 log_2 |8x – 56|$ or $log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?
$y+12 = log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.
answered Jan 30 at 18:29
VasyaVasya
4,2771618
$endgroup$
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
add a comment |
$begingroup$
The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.
You can always check your work by plugging your answer back into the original equation you were trying to solve.
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
You had a good start. $12 = log_2 (8x – 56)^4$ can be rewritten as $12 = 4 log_2 |8x – 56|$ or $log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?
$y+12 = log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.
answered Jan 30 at 18:29
VasyaVasya
4,2771618
$endgroup$
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
add a comment |
$begingroup$
You had a good start. $12 = log_2 (8x – 56)^4$ can be rewritten as $12 = 4 log_2 |8x – 56|$ or $log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?
$y+12 = log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.
answered Jan 30 at 18:29
VasyaVasya
4,2771618
$endgroup$
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
add a comment |
$begingroup$
You had a good start. $12 = log_2 (8x – 56)^4$ can be rewritten as $12 = 4 log_2 |8x – 56|$ or $log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?
$y+12 = log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.
answered Jan 30 at 18:29
VasyaVasya
4,2771618
$endgroup$
You had a good start. $12 = log_2 (8x – 56)^4$ can be rewritten as $12 = 4 log_2 |8x – 56|$ or $log_2 |8x – 56|=3$ which leads to $|8x-56|=8$. Can you finish?
$y+12 = log_2 (8x – 56)^4$ can be simplified to $2^{y/4}=|x-7|$.
answered Jan 30 at 18:29
VasyaVasya
4,2771618
edited Jan 31 at 3:10
answered Jan 30 at 18:29
VasyaVasya
4,2771618
answered Jan 30 at 18:29
VasyaVasya
4,2771618
answered Jan 30 at 18:29
VasyaVasya
4,2771618
4,2771618
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
add a comment |
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
I can finish. Thank you! One question, instead of |8x−56| = 8, is it supposed to be |8x−56| = y + 8 since we don't know the value of y. Y is not stated to be as 0.
$endgroup$
– moe
Jan 30 at 23:43
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
$begingroup$
Also, I forgot to mention that it says to simplify. would that make any changes?
$endgroup$
– moe
Jan 30 at 23:50
add a comment |
$begingroup$
The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.
You can always check your work by plugging your answer back into the original equation you were trying to solve.
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
add a comment |
$begingroup$
The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.
You can always check your work by plugging your answer back into the original equation you were trying to solve.
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
add a comment |
$begingroup$
The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.
You can always check your work by plugging your answer back into the original equation you were trying to solve.
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
The separation you did in step 2 is not valid. Log does not distribute over subtraction. Also (though your answer is way off by then), in step 7 you divided when you meant to multiply.
You can always check your work by plugging your answer back into the original equation you were trying to solve.
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
edited Jan 30 at 18:09
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 18:02
Nathaniel MayerNathaniel Mayer
1,863516
1,863516
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
add a comment |
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
For step one, (8x-56)^16, so I brought the 1/4 to the 16 which equals 4. Since 16/4 equals 4. In this case, the equation only equals to y. So it isn't like 4=4 its the y = 1/4 log2 (8x – 56)^16 – 12. Do you understand what I mean?
$endgroup$
– moe
Jan 30 at 18:08
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
Ah, sorry. I didn't notice the 16. Scrap that bit, look at step 2.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:09
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
No worries. So how do I solve the equation if I can't distribute the equation over subtraction? Do you know how? Thanks.
$endgroup$
– moe
Jan 30 at 18:13
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
$begingroup$
See Vasya's answer. At some point you need to apply the function $2^x$ to both sides.
$endgroup$
– Nathaniel Mayer
Jan 30 at 18:45
add a comment |
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$begingroup$
To be clear: Are you solving the equation $$y = frac14 log_2 (8x-56)^{16}-12$$ for $x$ when $y=0$? (BTW: One way to check your answer is to substitute it back into the original equation and see if you get what you expect.)
$endgroup$
– Blue
Jan 30 at 18:33
$begingroup$
This would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 30 at 19:02
$begingroup$
Hello Blue, the question just asks to simplify the equation.
$endgroup$
– moe
Jan 30 at 23:47