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Write the log equation as an exponential equation. You do not need to solve for x. log_((x+7))(3x-5)=2 Answer:

Write the log equation as an exponential equation. You do not need to solve for x \mathrm{x} .\newlinelog⁡(x+7)(3x−5)=2 \log _{(x+7)}(3 x-5)=2 \newlineAnswer:

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Full solution

Q. Write the log equation as an exponential equation. You do not need to solve for x \mathrm{x} .\newlinelog⁡(x+7)(3x−5)=2 \log _{(x+7)}(3 x-5)=2 \newlineAnswer:
  1. Identify Base, Exponent, Result: Identify the base, exponent, and result in the logarithmic equation.\newlineIn the equation log⁡(x+7)(3x−5)=2\log_{(x+7)}(3x-5)=2, the base is (x+7)(x+7), the exponent is 22, and the result is (3x−5)(3x-5).
  2. Rewrite in Exponential Form: Rewrite the logarithmic equation in exponential form.\newlineThe general form of a logarithm is log⁡base(result)=exponent\log_{\text{base}}(\text{result}) = \text{exponent}, which can be rewritten in exponential form as baseexponent=result\text{base}^{\text{exponent}} = \text{result}.\newlineTherefore, log⁡(x+7)(3x−5)=2\log_{(x+7)}(3x-5)=2 can be rewritten as $(x+\(7\))^\(2\) = \(3\)x\(-5\).

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