Mixing
Isolating for x in $log_x(3sqrt{x}) = k$
$begingroup$
I'm having trouble isolating for $x$ in $log_x(3sqrt{x}) = k$
I've tried various things. Here is what I ended up with:
$x^{k - frac{1}{2}} = 3$
I don't know how to proceed. I keep getting stuck. Can anyone help?
logarithms
asked Jan 17 at 16:15
898989898989
304
$endgroup$
add a comment |
$begingroup$
I'm having trouble isolating for $x$ in $log_x(3sqrt{x}) = k$
I've tried various things. Here is what I ended up with:
$x^{k - frac{1}{2}} = 3$
I don't know how to proceed. I keep getting stuck. Can anyone help?
logarithms
asked Jan 17 at 16:15
898989898989
304
$endgroup$
2
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
2
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10
add a comment |
$begingroup$
I'm having trouble isolating for $x$ in $log_x(3sqrt{x}) = k$
I've tried various things. Here is what I ended up with:
$x^{k - frac{1}{2}} = 3$
I don't know how to proceed. I keep getting stuck. Can anyone help?
logarithms
asked Jan 17 at 16:15
898989898989
304
$endgroup$
I'm having trouble isolating for $x$ in $log_x(3sqrt{x}) = k$
I've tried various things. Here is what I ended up with:
$x^{k - frac{1}{2}} = 3$
I don't know how to proceed. I keep getting stuck. Can anyone help?
logarithms
logarithms
asked Jan 17 at 16:15
898989898989
304
asked Jan 17 at 16:15
898989898989
304
edited Jan 17 at 21:37
898989
asked Jan 17 at 16:15
898989898989
304
asked Jan 17 at 16:15
898989898989
304
asked Jan 17 at 16:15
898989898989
304
304
2
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
2
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10
add a comment |
2
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
2
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10
2
2
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
2
2
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would have written the power as its definition, ie $x^{k-frac12}=e^{(k-frac12)ln(x)}$ and then take the neperian log on both sides, and get $(k-frac12) ln(x)=ln(3)$. Once you're here, it's quite easy to isolate $x$...
But every logarithm seems to work, so it's up to you !
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
$endgroup$
add a comment |
$begingroup$
May be
$(k-frac{1}{2})logx = log3$
and $x = 10^{left(dfrac{log(3)}{k-frac{1}{2}}right)}$
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would have written the power as its definition, ie $x^{k-frac12}=e^{(k-frac12)ln(x)}$ and then take the neperian log on both sides, and get $(k-frac12) ln(x)=ln(3)$. Once you're here, it's quite easy to isolate $x$...
But every logarithm seems to work, so it's up to you !
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
$endgroup$
add a comment |
$begingroup$
I would have written the power as its definition, ie $x^{k-frac12}=e^{(k-frac12)ln(x)}$ and then take the neperian log on both sides, and get $(k-frac12) ln(x)=ln(3)$. Once you're here, it's quite easy to isolate $x$...
But every logarithm seems to work, so it's up to you !
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
$endgroup$
add a comment |
$begingroup$
I would have written the power as its definition, ie $x^{k-frac12}=e^{(k-frac12)ln(x)}$ and then take the neperian log on both sides, and get $(k-frac12) ln(x)=ln(3)$. Once you're here, it's quite easy to isolate $x$...
But every logarithm seems to work, so it's up to you !
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
$endgroup$
I would have written the power as its definition, ie $x^{k-frac12}=e^{(k-frac12)ln(x)}$ and then take the neperian log on both sides, and get $(k-frac12) ln(x)=ln(3)$. Once you're here, it's quite easy to isolate $x$...
But every logarithm seems to work, so it's up to you !
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
answered Jan 17 at 16:33
Thomas PrévostThomas Prévost
1796
1796
add a comment |
add a comment |
$begingroup$
May be
$(k-frac{1}{2})logx = log3$
and $x = 10^{left(dfrac{log(3)}{k-frac{1}{2}}right)}$
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
add a comment |
$begingroup$
May be
$(k-frac{1}{2})logx = log3$
and $x = 10^{left(dfrac{log(3)}{k-frac{1}{2}}right)}$
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
add a comment |
$begingroup$
May be
$(k-frac{1}{2})logx = log3$
and $x = 10^{left(dfrac{log(3)}{k-frac{1}{2}}right)}$
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
$endgroup$
May be
$(k-frac{1}{2})logx = log3$
and $x = 10^{left(dfrac{log(3)}{k-frac{1}{2}}right)}$
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
answered Jan 17 at 16:23
Satish RamanathanSatish Ramanathan
10k31323
10k31323
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
add a comment |
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
In your answer, you're using base 10 logarithm. Here, it's a base $x$ logarithm...
$endgroup$
– Thomas Prévost
Jan 17 at 16:24
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
@ThomasPrévost He is continuing from the last step , and is taking log base 10 on either side
$endgroup$
– Sinπ
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
The expression stands correct by definition
$endgroup$
– Satish Ramanathan
Jan 17 at 16:25
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@RahulRamachandran my bad, I didn't see you took the log at the begining... So it's correct, I apologize
$endgroup$
– Thomas Prévost
Jan 17 at 16:28
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
$begingroup$
@ThomasPrévost, No worries!!
$endgroup$
– Satish Ramanathan
Jan 17 at 16:29
add a comment |
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2
$begingroup$
You’re only a step away, raise both sides to the reciprocal of $k-frac{1}{2}$.
$endgroup$
– KM101
Jan 17 at 16:20
$begingroup$
You have done your best. It seems correct and I do not know how else you could simplify. x seems to be the function for some value of k
$endgroup$
– Satish Ramanathan
Jan 17 at 16:20
2
$begingroup$
$$log_x(3sqrt x)=kimplies x^k=3x^{1/2}implies x^{k-1/2}=3implies x=3^{1/(k-1/2)}$$
$endgroup$
– Mark Viola
Jan 17 at 17:10