How do you solve $(0.3)^(1 + x) = 1.7^(2x - 1)$ ?

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Answer 1 · 2016-04-13T16:46:00

$x=(log 1.7+log 0.3) /(2 log 1.7 - log 0.3)=-0.2973$ , nearly..

$log a^m=m log a$
Here, $(0.3)^(1+x)=(1.7)^(2x-1)$
Equating logarithms.
$(1+x) log 0.3=(2x-1) log 1.7$

Solving for x.
$x=(log 1.7+log 0.3) /(2 log 1.7 - log 0.3)$

How do you solve (0.3)^(1 + x) = 1.7^(2x - 1)(0.3)^(1 + x) = 1.7^(2x - 1)?