J jimmy1 Jul 2010 1 0 Jul 29, 2010 #1 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?

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?

A Ackbeet MHF Hall of Honor Jun 2010 6,318 2,433 CT, USA Jul 29, 2010 #2 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.

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.

B bobak Oct 2007 591 254 London / Cambridge Jul 29, 2010 #3 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.

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 Jan 2010 278 138 Jul 29, 2010 #4 jimmy1 said: 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? Click to expand... \(\displaystyle \begin{aligned} 7^{-4x} &= 2^{1 + 3x} \\ -4x \ln 7 &= (1+3x) \ln 2 \\ -4x \ln 7 &= \ln 2 + 3x \ln 2 \\ 3x \ln 2 + 4x \ln 7 &= -\ln 2 \\ x(3\ln 2 + 4\ln 7) &= -\ln 2 ... \\ \end{aligned}\) Can you finish? Reactions: jkrowling

jimmy1 said: 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? Click to expand... \(\displaystyle \begin{aligned} 7^{-4x} &= 2^{1 + 3x} \\ -4x \ln 7 &= (1+3x) \ln 2 \\ -4x \ln 7 &= \ln 2 + 3x \ln 2 \\ 3x \ln 2 + 4x \ln 7 &= -\ln 2 \\ x(3\ln 2 + 4\ln 7) &= -\ln 2 ... \\ \end{aligned}\) Can you finish?