Certainly! Let's solve the exponential equation [tex]\(50^x = 17\)[/tex] step by step.
To solve for [tex]\(x\)[/tex], we'll use logarithms and the change of base formula. The change of base formula states that for any positive numbers [tex]\(b\)[/tex] and [tex]\(y\)[/tex] (with [tex]\(b \neq 1\)[/tex]):
[tex]\[ \log_b y = \frac{\log y}{\log b} \][/tex]
Here, [tex]\(b\)[/tex] is the base of the logarithm, and [tex]\(y\)[/tex] is the argument.
1. Start with the given equation:
[tex]\[ 50^x = 17 \][/tex]
2. Take the logarithm of both sides. You can use any logarithm base (common logarithm, natural logarithm, etc.), but we will use the natural logarithm (denoted [tex]\(\ln\)[/tex]) for this solution:
[tex]\[ \ln(50^x) = \ln(17) \][/tex]
3. Use the logarithm power rule, which states that [tex]\(\ln(a^b) = b \cdot \ln(a)\)[/tex]:
[tex]\[ x \cdot \ln(50) = \ln(17) \][/tex]
4. Solve for [tex]\(x\)[/tex] by isolating it on one side of the equation. Divide both sides by [tex]\(\ln(50)\)[/tex]:
[tex]\[ x = \frac{\ln(17)}{\ln(50)} \][/tex]
5. Calculate the logarithms. Using a calculator:
[tex]\[\ln(17) \approx 2.833213344 \][/tex]
[tex]\[\ln(50) \approx 3.912023005 \][/tex]
6. Divide the results:
[tex]\[ x \approx \frac{2.833213344}{3.912023005} \][/tex]
[tex]\[ x \approx 0.7242322808748767 \][/tex]
So, the value of [tex]\(x\)[/tex] that satisfies the equation [tex]\(50^x = 17\)[/tex] is approximately [tex]\(0.7242322808748767\)[/tex].
