# Logarithmic Help

• Jul 17th 2008, 04:50 PM
Maga22
Logarithmic Help
I tried the following problem:

log (log x) = 1

I need to know if I did it right.

Problem 1:

log(log x) = 1

log x = e^1
log x = e

The second problem:

3^log x = 3x

I attempted but got stuck. Any help will be much appreciated.

3^log x = 3x

log(3^log x) = log 3x
log3logx = log3+logx
• Jul 17th 2008, 05:06 PM
Jhevon
Quote:

Originally Posted by Maga22
I tried the following problem:

log (log x) = 1

I need to know if I did it right.

Problem 1:

log(log x) = 1

log x = e^1
log x = e

yes (Clapping)

Quote:

The second problem:

3^log x = 3x

I attempted but got stuck. Any help will be much appreciated.

3^log x = 3x

log(3^log x) = log 3x
log3logx = log3+logx
nice start. now just treat log(x) as a variable and solve for it. what if you have (log3)y = log3 + y. could you solve for y? sure you could. here, of course, y = logx

we get $y = \frac {\log 3}{\log 3 - 1}$

so, $\log x = \frac {\log 3}{\log 3 - 1}$

can you finish?
• Jul 18th 2008, 01:08 AM
Maga22
Logarithmic Help
Thanks for the feed back on problem one. I am still stuck on the second problem. This is as far as I got:

Problem: 3^logx=3x

log(3^logx)=log3x
log 3 log x=log 3+log x
(logx)^2=log 3+log x
(logx)^2-log x-log 3=0
(logx)^2-logx ??

Can you help please? I'm a little confused with the sample provided. I attempted again, but not sure if I am on the right track. I have several more problems like this and I need to have some sort of sample to go off of in order to start and complete the others. Thanks.
• Jul 18th 2008, 01:16 AM
flyingsquirrel
Hi
Quote:

Originally Posted by Maga22
Thanks for the feed back on problem one. I am still stuck on the second problem. This is as far as I got:

Problem: 3^logx=3x

log(3^logx)=log3x
log 3 log x=log 3+log x
(logx)^2=log 3+log x

Where does $(\log x)^{\color{red}2}$ come from ?

From $\log 3 \log x =\log 3+\log x$ we get, subtracting $\log x$ to both sides, $\log 3 \log x -\log x=\log 3$

Factor : $(\log 3 -1)\log x =\log 3$

Can you take it from here ?
• Jul 18th 2008, 01:17 AM
Chris L T521
Quote:

Originally Posted by Maga22
I tried the following problem:

log (log x) = 1

I need to know if I did it right.

Problem 1:

log(log x) = 1

log x = e^1
log x = e

Quote:

Originally Posted by Jhevon
yes (Clapping)

Sorry man, but I disagree. $e^{\log(\log(x))}\neq\log(x)$ However, $10^{\log(\log(x))}=\log(x)$ since $\log(x)=\log_{\color{red}10}(x)$. If the problem had contained $\ln(x)$ or $\log_{\color{red}e}(x)$, then the answer would be $x=e^e$...

Thus,

$10^{\log(\log(x))}=10^{1}$

$\implies\log(x)=10$

$\implies \color{red}\boxed{x=10^{10}}$

That should be the correct answer.

Quote:

Originally Posted by Maga22
The second problem:

3^log x = 3x

I attempted but got stuck. Any help will be much appreciated.

3^log x = 3x

log(3^log x) = log 3x
log3logx = log3+logx

Quote:

Originally Posted by Jhevon
nice start. now just treat log(x) as a variable and solve for it. what if you have (log3)y = log3 + y. could you solve for y? sure you could. here, of course, y = logx

we get $y = \frac {\log 3}{\log 3 - 1}$

so, $\log x = \frac {\log 3}{\log 3 - 1}$

can you finish?

Good. I have no problems with this.

--Chris
• Jul 18th 2008, 01:29 AM
Moo
Hello Chris,

I think it depends on what you intend by $\log$. I've seen several websites in English where $\log$ is used for base e. Though most of them use it for base 10...
• Jul 18th 2008, 01:29 AM
Maga22
Quote:

Originally Posted by flyingsquirrel
Hi

Where does $(\log x)^{\color{red}2}$ come from ?

From $\log 3 \log x =\log 3+\log x$ we get, subtracting $\log x$ to both sides, $\log 3 \log x -\log x=\log 3$

Factor : $(\log 3 -1)\log x =\log 3$

Can you take it from here ?

I divided both sides by log 3-1 and got this

http://www.mathhelpforum.com/math-he...f97ce496-1.gif

This where I got stuck earlier. Can someone please assist?
• Jul 18th 2008, 01:33 AM
Chris L T521
Quote:

Originally Posted by Moo
Hello Chris,

I think it depends on what you intend by $\log$. I've seen several websites in English where $\log$ is used for base e. Though most of them use it for base 10...

Moo, you have a point. In America, I was taught that $\log(x)=\log_{10} (x)$, and that $\ln(x)=\log_e(x)$. However, as you mentioned, there are other ways it can be interpreted [i.e. $\log(x)=\log_e(x)$]. I'll keep that in mind as other problems like this pop up.

--Chris
• Jul 18th 2008, 01:49 AM
flyingsquirrel
Quote:

Originally Posted by Maga22
I divided both sides by log 3-1 and got this

http://www.mathhelpforum.com/math-he...f97ce496-1.gif

This where I got stuck earlier. Can someone please assist?

If $\log$ is the natural logarithm (i.e. $\exp(\log x )=x$) then $\log x=\frac{\log 3}{\log 3-1}\implies \underbrace{\exp (\log x)}_{x}=\exp\left( \frac{\log 3}{\log 3-1}\right)\implies x=\exp\left( \frac{\log 3}{\log 3-1}\right)$

If $\log$ is the base 10 logarithm then $10^{\log x}=x$ and, exactly as in the previous case, we get $x=10^{\frac{\log 3}{\log 3-1}}$