Question

find the value of x if 3^x+64=2^6+root3^8

Answer 1

Answer:

4

$2^6=64$

64 will cancel out.

$\sqrt[b]{a} =a^{\frac{1}{b} }$ for $a>0$ and positive integer $b\geq 2$.

$(\sqrt[2]{3} )^8=3^{\frac{1}{2} \times 8}=3^4$

$3^x=3^4$

By the logarithm, with the base 3

∴ $x=4$

Answer 2

Answer:

3^x + 64 = 2^6 + √3^8

3^x + 64 = 64 + (√3^2)^4

Subtract 64 on both sides,

3^x + 64 - 64 = 64 - 64 + (√3^2)^4

3^x = (√3^2)^4

3^x = 3^4

Since, Base is same, therefore, the exponents will also be same.

Therefore, x = 4

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