CUET UG 2025 — General Test Quantitative Reasoning

Given: $\log_8 x = \frac{2}{3}$log8​x=32​

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 $\log_b x = y$logb​x=y, then $b^y = x$by=x

For this problem:

• Base $(b) = 8$(b)=8

• Result $(y) = \frac{2}{3}$(y)=32​

• Unknown $(x) = ?$(x)=?

Therefore:

$$x = 8^{\frac{2}{3}}$$x=832​

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The fractional exponent $\frac{2}{3}$32​ means:

• The denominator $(3)$(3) = take the cube root

• The numerator $(2)$(2) = square the result

So:

$$8^{\frac{2}{3}} = (\sqrt[3]{8})^2$$832​=(38​)2

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Find the cube root of $8$8:

$\sqrt[3]{8} = 2$38​=2 (because $2 \times 2 \times 2 = 8$2×2×2=8)

Square that result:

$$(2)^2 = 4$$(2)2=4

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Therefore, $x = 4$x=4