# Solve Logarithm

• Oct 28th 2008, 09:27 AM
ellimist121
Solve Logarithm
I'm having the same kind of problems on all of the following problems: I cannot seem to isolate X. Help if you can!

1. 5^(x+1)=e^x
2. 3^(x+1)=(9^x)(27^(x+1))
3. 10^(2x+1)=4^(x+1)
4. 5^(3x-2)=3^(x+1)
5. *Log Base is 2* Log(x+4) + Log(x+2)=Log3

Much appreciated!
• Oct 28th 2008, 10:20 AM
ThePerfectHacker
Quote:

Originally Posted by ellimist121
1. 5^(x+1)=e^x

Take $\displaystyle \ln$ of both sides to get, $\displaystyle \ln 5^{x+1} = \ln e^x$. Using identities we get $\displaystyle (x+1) \ln 5 = x \ln e$. But $\displaystyle \ln e = 1$. Thus, we are left with $\displaystyle (x+1)\ln 5 = x \implies x \ln 5 + \ln 5 = x$. Rewrite as $\displaystyle x\ln 5 - x = - \ln 5$. Factor, $\displaystyle x(\ln 5 - 1) = - \ln 5$. And this gives $\displaystyle x = - \tfrac{\ln 5}{\ln 5 -1} = \tfrac{\ln 5}{1-\ln 5}$

Quote:

2. 3^(x+1)=(9^x)(27^(x+1))
This is,
$\displaystyle 3^x = 3^{2x} \cdot 3^{3(x+1)} \implies 3^x = 3^{2x + 3(x+1)} \implies x = 2x + 3(x+1)$
Now solve for $\displaystyle x$

Try doing the other problems.
• Oct 28th 2008, 12:21 PM
ellimist121
Thanks, but I'm still having trouble with the last one I have gotten to this:

Log(((x+4)(x+2))/3)=0

Also, with this one:

(5^2x)-7(5^x)+12=0.
• Oct 28th 2008, 12:44 PM
11rdc11
Quote:

Originally Posted by ellimist121
Thanks, but I'm still having trouble with the last one I have gotten to this:

Log(((x+4)(x+2))/3)=0

Also, with this one:

(5^2x)-7(5^x)+12=0.

$\displaystyle 2^0 = \frac{x^2+6x+8}{3}$