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?
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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.
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.
Originally Posted by 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? $\displaystyle \begin{aligned} 7^{-4x} &= 2^{1 + 3x} \\ -4x \ln 7 &= (1+3x) \ln 2 \\ -4x \ln 7 &= \ln 2 + 3x \ln 2 \\ 3x \ln 2 + 4x \ln 7 &= -\ln 2 \\ x(3\ln 2 + 4\ln 7) &= -\ln 2 ... \\ \end{aligned}$ Can you finish?
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