Mixing
Let 𝐿(𝑥)=log𝑎(𝑥) where we don't know the base 𝑎.
$begingroup$
Let 𝐿(𝑥)=log𝑎(𝑥) where we don't know the base 𝑎.
However, we do know that
𝐿(2)=0.37031
and
𝐿(3)=0.58692
Use this information to compute
𝐿(4)=
𝐿(𝑎^2)=
𝐿(𝑎^3)=
𝐿(6^5)=
I tried to do 0.37031=log𝑎2 and 0.58692=log𝑎3 and convert both to exponential form but I'm not sure how to find the base 𝑎 value. Any tips on how to properly start this question would be appreciated!
logarithms
asked Jan 19 at 23:41
mfmmfm
11
$endgroup$
add a comment |
$begingroup$
Let 𝐿(𝑥)=log𝑎(𝑥) where we don't know the base 𝑎.
However, we do know that
𝐿(2)=0.37031
and
𝐿(3)=0.58692
Use this information to compute
𝐿(4)=
𝐿(𝑎^2)=
𝐿(𝑎^3)=
𝐿(6^5)=
I tried to do 0.37031=log𝑎2 and 0.58692=log𝑎3 and convert both to exponential form but I'm not sure how to find the base 𝑎 value. Any tips on how to properly start this question would be appreciated!
logarithms
asked Jan 19 at 23:41
mfmmfm
11
$endgroup$
$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12
add a comment |
$begingroup$
Let 𝐿(𝑥)=log𝑎(𝑥) where we don't know the base 𝑎.
However, we do know that
𝐿(2)=0.37031
and
𝐿(3)=0.58692
Use this information to compute
𝐿(4)=
𝐿(𝑎^2)=
𝐿(𝑎^3)=
𝐿(6^5)=
I tried to do 0.37031=log𝑎2 and 0.58692=log𝑎3 and convert both to exponential form but I'm not sure how to find the base 𝑎 value. Any tips on how to properly start this question would be appreciated!
logarithms
asked Jan 19 at 23:41
mfmmfm
11
$endgroup$
Let 𝐿(𝑥)=log𝑎(𝑥) where we don't know the base 𝑎.
However, we do know that
𝐿(2)=0.37031
and
𝐿(3)=0.58692
Use this information to compute
𝐿(4)=
𝐿(𝑎^2)=
𝐿(𝑎^3)=
𝐿(6^5)=
I tried to do 0.37031=log𝑎2 and 0.58692=log𝑎3 and convert both to exponential form but I'm not sure how to find the base 𝑎 value. Any tips on how to properly start this question would be appreciated!
logarithms
logarithms
asked Jan 19 at 23:41
mfmmfm
11
asked Jan 19 at 23:41
mfmmfm
11
edited Jan 19 at 23:51
mfm
asked Jan 19 at 23:41
mfmmfm
11
asked Jan 19 at 23:41
mfmmfm
11
asked Jan 19 at 23:41
mfmmfm
11
11
$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12
add a comment |
$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12
$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12
add a comment |
1 Answer
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$begingroup$
Just do it.
You know $log_a b^m = mlog_a b$ so you can use this to solve $L(4) = log_a 2^2$.
And you know $log_a a^k = k$ which you can use to solve the next two.
And you know $log_a bc = log_a b + log_a c$ which you can use to solve the last one.
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
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$begingroup$
Just do it.
You know $log_a b^m = mlog_a b$ so you can use this to solve $L(4) = log_a 2^2$.
And you know $log_a a^k = k$ which you can use to solve the next two.
And you know $log_a bc = log_a b + log_a c$ which you can use to solve the last one.
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Just do it.
You know $log_a b^m = mlog_a b$ so you can use this to solve $L(4) = log_a 2^2$.
And you know $log_a a^k = k$ which you can use to solve the next two.
And you know $log_a bc = log_a b + log_a c$ which you can use to solve the last one.
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
$endgroup$
add a comment |
$begingroup$
Just do it.
You know $log_a b^m = mlog_a b$ so you can use this to solve $L(4) = log_a 2^2$.
And you know $log_a a^k = k$ which you can use to solve the next two.
And you know $log_a bc = log_a b + log_a c$ which you can use to solve the last one.
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
$endgroup$
Just do it.
You know $log_a b^m = mlog_a b$ so you can use this to solve $L(4) = log_a 2^2$.
And you know $log_a a^k = k$ which you can use to solve the next two.
And you know $log_a bc = log_a b + log_a c$ which you can use to solve the last one.
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
answered Jan 20 at 0:05
fleabloodfleablood
71.7k22686
71.7k22686
add a comment |
add a comment |
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$begingroup$
Thanks saulspatz!
$endgroup$
– mfm
Jan 19 at 23:52
$begingroup$
Please review the properties of logarithms. Also the fact that $log_a a=1$
$endgroup$
– GReyes
Jan 19 at 23:53
$begingroup$
A more general version of the equation which GReyes provided is $log_a{a^n} = n$ for all $n$. You can use this solve $2$ of the values you are trying to compute.
$endgroup$
– John Omielan
Jan 19 at 23:59
$begingroup$
And $log_a mn = log_a m + log_a n$ will solve all the rest.
$endgroup$
– fleablood
Jan 20 at 0:06
$begingroup$
It would be very mean of me to actually answer your question. But I am mean so I will. If $log_a 2 = 0.37031$ then $a^{0.37031} =2$ so $ln a^{0.3701} = ln 2$ so $0.3701 ln a = ln 2$ so $ln a = frac {ln 2}{0.3701}$ so $a = e^{frac {ln 2}{0.3701}} = 2^{frac 1{0.3701}} = 2^{frac {10000}{3701}}$
$endgroup$
– fleablood
Jan 20 at 0:12