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(05.02 MC)

Explain how to solve [tex]$5^{x-2}=8$[/tex] using the change of base formula [tex]\log_b y = \frac{\log y}{\log b}[/tex]. Include the solution for [tex]x[/tex] in your answer. Round your answer to the nearest thousandth.


Sagot :

To solve the equation [tex]\(5^{x-2} = 8\)[/tex] using logarithms and the change of base formula, follow these steps:

1. Take the logarithm of both sides of the equation: This allows us to bring the exponent down and make use of the logarithmic properties.
[tex]\[ \log(5^{x-2}) = \log(8) \][/tex]

2. Use the power rule of logarithms: The power rule states that [tex]\(\log(a^b) = b \log(a)\)[/tex]. Apply this to the left-hand side of the equation.
[tex]\[ (x-2) \log(5) = \log(8) \][/tex]

3. Isolate the term involving [tex]\(x\)[/tex]: To solve for [tex]\(x\)[/tex], we need to isolate it. Start by dividing both sides of the equation by [tex]\(\log(5)\)[/tex].
[tex]\[ x - 2 = \frac{\log(8)}{\log(5)} \][/tex]

4. Solve for [tex]\(x\)[/tex]: Add 2 to both sides to isolate [tex]\(x\)[/tex].
[tex]\[ x = \frac{\log(8)}{\log(5)} + 2 \][/tex]

5. Calculate the logarithms and the final value of [tex]\(x\)[/tex]:
- [tex]\(\log(8) \approx 2.079\)[/tex]
- [tex]\(\log(5) \approx 1.609\)[/tex]
- So,
[tex]\[ x = \frac{2.079}{1.609} + 2 \approx 1.292 + 2 = 3.292 \][/tex]

6. Round the answer to the nearest thousandth:
[tex]\[ x \approx 3.292 \][/tex]

Therefore, the solution to the equation [tex]\(5^{x-2} = 8\)[/tex] is [tex]\(x \approx 3.292\)[/tex] when rounded to the nearest thousandth.
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