# Thread: Logarithmic Help

1. ## 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

ANSWER: x = e^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

2. 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

ANSWER: x = e^e
yes

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?

3. ## 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.

4. Hi
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 ?

5. 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

ANSWER: x = e^e
Originally Posted by Jhevon
yes
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.

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
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

6. 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...

7. 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

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

8. 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

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

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 $x=10^{\frac{\log 3}{\log 3-1}}$