if (10^{log x}=\frac{1}{10^{32}}), what is the value of (x)? (1 point)\n\n(10^{-32})\n\n32\n\n(10^{32})\n\n…

Explanation:

Step1: Use logarithm - exponential property

Recall that (a^{\log_a x}=x). For the left - hand side (10^{\log x}), since the base of the exponential is 10 and the base of the logarithm is also 10 (common logarithm, base 10), then (10^{\log x}=x).

Step2: Rewrite the right - hand side

We know that (\frac{1}{10^{32}} = 10^{-32}) according to the rule (\frac{1}{a^n}=a^{-n}) where (a = 10) and (n = 32).

Step3: Solve for x

Since (10^{\log x}=x) and (10^{\log x}=\frac{1}{10^{32}}=10^{-32}), then (x = 10^{-32}).

Answer:

(10^{-32})