If $\log_8 x = 2/3$, then the value of X is:

Given: $\log_8 x = \frac{2}{3}$

A logarithm asks: "What power do I raise the base to, to get x?"

So this is saying: "2/3 is the power I raise 8 to, to get x"

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The basic rule is: If $\log_b x = y$, then $b^y = x$

For this problem:

• Base $(b) = 8$
• Result $(y) = \frac{2}{3}$
• Unknown $(x) = ?$

Therefore:

$x = 8^{\frac{2}{3}}$

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The fractional exponent $\frac{2}{3}$ means:

• The denominator $(3)$ = take the cube root
• The numerator $(2)$ = square the result

So:

$8^{\frac{2}{3}} = (\sqrt[3]{8})^2$

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Find the cube root of $8$:

$\sqrt[3]{8} = 2$ (because $2 \times 2 \times 2 = 8$)

Square that result:

$(2)^2 = 4$

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Therefore, $x = 4$