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

#### jimmy1

Help:

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

How do I simplify?

#### Ackbeet

MHF Hall of Honor
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

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

Help:

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

How do I simplify?
\displaystyle \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?

jkrowling