X in exponent on both sides, can't make bases the same

**jimmy1** · July 29th 2010, 12:36 PM

Help:

$7^-^4^x = 2^1^+^3^x$
I should take the ln of both sides, right...?
$-4x ln 7 = (1+3x) ln 2$

How do I simplify?

**Ackbeet** · July 29th 2010, 12:43 PM

I would distribute out the RHS first. Your goal, as always is to get the terms with x's on one side, and all the other terms on the other.

**bobak** · July 29th 2010, 12:43 PM

You don't want to simplify, you want to solve the equation. Step back and look at what you have written, remember that ln7 and ln2 are just numbers.

**eumyang** · July 29th 2010, 12:48 PM

Originally Posted by jimmy1

Help:

$7^-^4^x = 2^1^+^3^x$
I should take the ln of both sides, right...?
$-4x ln 7 = (1+3x) ln 2$

How do I simplify?

$\begin{aligned} 7^{-4x} &= 2^{1 + 3x} \\ -4x \ln 7 &= (1+3x) \ln 2 \\ -4x \ln 7 &= \ln 2 + 3x \ln 2 \\ 3x \ln 2 + 4x \ln 7 &= -\ln 2 \\ x(3\ln 2 + 4\ln 7) &= -\ln 2 ... \\ \end{aligned}$

Can you finish?

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