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

Jul 2010
1
0
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
Jun 2010
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CT, USA
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.
 
Oct 2007
591
254
London / Cambridge
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.
 
Jan 2010
278
138
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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