How do you solve ${log}_{8} (x + 1) = \frac{2}{3}$ $\log _ { 8} ( x + 1) = \frac { 2} { 3}$ ?

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How do you solve ${log}_{8} (x + 1) = \frac{2}{3}$ $\log _ { 8} ( x + 1) = \frac { 2} { 3}$ ?

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Answer 1 · 2017-04-27T05:11:43

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Explanation:

Given
${log}_{8} (x + 1) = \frac{2}{3}$ $log_8(x+1)=2/3$
By Logarithmic property ${log}_{a} b = c$ $color(blue)(log_ab=c)$ can be written as $a^{c} = b$ $color(blue)(a^c=b)$
$8^{\frac{2}{3}} = x + 1$ $8^(2/3)=x+1$
then
$x = 8^{\frac{2}{3}} - 1$ $x=8^(2/3)-1$
$x = 4 - 1$ $x=4-1$
$x = 3$ $x=3$

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How do you solve ${log}_{8} (x + 1) = \frac{2}{3}$ $\log _ { 8} ( x + 1) = \frac { 2} { 3}$ ?

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How do you solve ${log}_{8} (x + 1) = \frac{2}{3}$ $\log _ { 8} ( x + 1) = \frac { 2} { 3}$ ?