Question

32^2/3 =3^-4 = log32 43 - 2log2 1 =

Answer 1

$32^{\frac{2}{3} } =\sqrt[3]{32^{2} } =\sqrt[3]{1024} =8\sqrt[3]{2} \\ \\ 3^{-4} =\frac{1}{3^{4} } =\frac{1}{81}$

㏒₃₂(43)-2㏒₂(1)=㏒₃₂(43)-2·0=㏒₃₂(43)-0=㏒₃₂(43)=㏒₂⁵(43)=(1/5)㏒₂(43)

Answer 2

Let's break down each expression step by step:

1. 32^(2/3): This means raising 32 to the power of 2/3. 32^(2/3) = (2^5)^(2/3) = 2^(5 * 2/3) = 2^(10/3)

2. 3^(-4): This means raising 3 to the power of -4. 3^(-4) = 1/(3^4) = 1/(3^2 * 3^2) = 1/(9 * 9) = 1/81

3. log32 43: This represents the logarithm of 43 with base 32. log32 43 = log(43)/log(32)

4. 2log2 1: This expression represents twice the logarithm of 1 with base 2. log2 1 = 0 (since any number raised to the power of 0 is 1) 2log2 1 = 2 * 0 = 0

Now let's substitute the values back into the equation:

32^(2/3) = 3^(-4) = log32 43 - 2log2 1

2^(10/3) = 1/81 = log(43)/log(32) - 0

At this point, we can solve for log(43)/log(32) by isolating the term:

2^(10/3) + 0 = log(43)/log(32)

Now, let's evaluate the left side of the equation:

2^(10/3) ≈ 18.924

Substituting this value back into the equation:

18.924 ≈ log(43)/log(32)

To find the exact values, you would need to use a calculator or software that can calculate logarithms.