How to find the x in this exponential function? Thank you!

Question

$2(7^x+2^(x-1))=7(2^x-7^(x-1))$

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Answer 1 · 2018-05-15T23:32:26

$log_3.5(2)$

Distributing the variables on each side gives us
$2*7^x+2*2^(x-1) = 7*2^x-7*7^(x-1)$

We know $a^(x-1)*a=a^x$ using basic exponent rules, so applying that rule gives us

$2*7^x+2^x=7*2^x-7^x$
Combine like terms
$3*7^x=6*2^x$
$7^x/2^x=2$

Using another exponent rule, we know $a^x/b^2=(a/b)^x$ , so
$(7/2)^x=2$
$3.5^x=2$

Using logarithms to solve, we arrive at

$x=log_3.5(2)$