How do you solve $8^{\sqrt{x}} = 4^{\sqrt[3]{x}}$ $8^sqrtx = 4^root3 x$ ?

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How do you solve $8^{\sqrt{x}} = 4^{\sqrt[3]{x}}$ $8^sqrtx = 4^root3 x$ ?

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Answer 1 · 2017-02-23T03:49:01

${0, \frac{64}{729}}$ ${0, 64/729}$

Explanation:

Rewrite in base $2$ $2$ .

${(2^{3})}^{\sqrt{x}} = {(2^{2})}^{\sqrt[3]{x}}$ $(2^3)^sqrt(x) = (2^2)^root(3)(x)$

$2^{3 \sqrt{x}} = 2^{2 \sqrt[3]{x}}$ $2^(3sqrt(x)) = 2^(2root(3)(x)$

We don't need the bases anymore.

$3 \sqrt{x} = 2 \sqrt[3]{x}$ $3sqrt(x) = 2root(3)(x)$

$3 x^{\frac{1}{2}} = 2 x^{\frac{1}{3}}$ $3x^(1/2) = 2x^(1/3)$

Put both sides to the 6th power to get rid of fractional exponents (we take the 6th power because the LCM of $2$ $2$ and $3$ $3$ is $6$ $6$ ).

${(3 x^{\frac{1}{2}})}^{6} = {(2 x^{\frac{1}{3}})}^{6}$ $(3x^(1/2))^6 = (2x^(1/3))^6$

$729 x^{3} = 64 x^{2}$ $729x^3 = 64x^2$

$729 x^{3} - 64 x^{2} = 0$ $729x^3 - 64x^2 = 0$

$x^{2} (729 x - 64) = 0$ $x^2(729x - 64) = 0$

$x = 0 and \frac{64}{729}$ $x = 0 and 64/729$

Hopefully this helps!

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