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

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

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Answer 1 · 2016-04-13T16:46:00

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

Explanation:

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

Solving for x.
$x = \frac{log 1.7 + log 0.3}{2 log 1.7 - log 0.3}$ $x=(log 1.7+log 0.3) /(2 log 1.7 - log 0.3)$

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

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Explanation:

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