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

• Jul 29th 2010, 12:36 PM
jimmy1
X in exponent on both sides, can't make bases the same
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?
• Jul 29th 2010, 12:43 PM
Ackbeet
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.
• Jul 29th 2010, 12:43 PM
bobak
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.
• Jul 29th 2010, 12:48 PM
eumyang
Quote:

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?