Solve Logarithm

**ellimist121** · Oct 28th 2008, 09:27 AM

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!

**ThePerfectHacker** · Oct 28th 2008, 10:20 AM

Originally Posted by ellimist121

1. 5^(x+1)=e^x

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

2. 3^(x+1)=(9^x)(27^(x+1))

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

Try doing the other problems.

**ellimist121** · Oct 28th 2008, 12:21 PM

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.

**11rdc11** · Oct 28th 2008, 12:44 PM

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.

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

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