Question

question 6 (1 point)
solve for x: \\(\log_x 16 = \frac{4}{5}\\)
\\(\bigcirc\\) a) 32
\\(\bigcirc\\) b) -1.3
\\(\bigcirc\\) c) 64
\\(\bigcirc\\) d) 0.96
question 7 (1 point)

Explanation:

Step1: Recall logarithm definition

If \(\log_{a}b = c\), then \(a^{c}=b\). Here, \(\log_{x}16=\frac{4}{3}\), so \(x^{\frac{4}{3}} = 16\).

Step2: Isolate x

Raise both sides to the power of \(\frac{3}{4}\) to solve for \(x\). Using the exponent rule \((a^{m})^{n}=a^{mn}\), we get \(x=(16)^{\frac{3}{4}}\).

Step3: Simplify 16^(3/4)

First, \(16 = 2^{4}\), so \((2^{4})^{\frac{3}{4}}=2^{4\times\frac{3}{4}}=2^{3}=8\)? Wait, no, wait, maybe I misread the original problem. Wait, the original problem: \(\log_{x}16=\frac{4}{3}\)? Wait, maybe the original problem is \(\log_{x}16=\frac{4}{3}\)? Wait, no, let's check again. Wait, if \(\log_{x}16=\frac{4}{3}\), then \(x^{4/3}=16\), so \(x = 16^{3/4}\). But \(16^{1/4}=2\), so \(16^{3/4}=(16^{1/4})^{3}=2^{3}=8\)? But 8 is not in the options. Wait, maybe the original problem is \(\log_{x}16=\frac{4}{5}\)? Wait, the user's image shows \(\log_{x}16=\frac{4}{5}\)? Wait, the user's question: "Solve for x: \(\log_{x}16=\frac{4}{5}\)"? Wait, the options are a)32, b)-1.3, c)64, d)0.96. Let's re - do it with \(\frac{4}{5}\).

Step1 (corrected): Recall logarithm definition

If \(\log_{a}b = c\), then \(a^{c}=b\). So for \(\log_{x}16=\frac{4}{5}\), we have \(x^{\frac{4}{5}}=16\).

Step2 (corrected): Isolate x

Raise both sides to the power of \(\frac{5}{4}\): \(x=(16)^{\frac{5}{4}}\).

Step3 (corrected): Simplify 16^(5/4)

Since \(16 = 2^{4}\), then \((2^{4})^{\frac{5}{4}}=2^{4\times\frac{5}{4}}=2^{5}=32\). Ah, that's option a. So the original problem must have \(\frac{4}{5}\) instead of \(\frac{4}{3}\) (maybe a typo in my initial reading). So with \(\log_{x}16=\frac{4}{5}\), we get \(x = 32\).

Answer:

a) 32