If 3^{x-2}=5 and log_{10} 2=0.30103, log_{10} | vedclass

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(C) Given $3^{x-2} = 5$.Taking $\log_{10}$ on both sides:$(x-2) \log_{10} 3 = \log_{10} 5$Since $\log_{10} 5 = \log_{10} (10/2) = \log_{10} 10 - \log_

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$3 \frac{22187}{47710}$

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(C) Given $3^{x-2} = 5$.Taking $\log_{10}$ on both sides:$(x-2) \log_{10} 3 = \log_{10} 5$Since $\log_{10} 5 = \log_{10} (10/2) = \log_{10} 10 - \log_{10} 2 = 1 - 0.30103 = 0.69897$.So,$(x-2) (0.4771) = 0.69897$.$x-2 = \frac{0.69897}{0.4771} = 1.465038...$$x = 2 + 1.465038 = 3.465038$.Calculating the fraction: $3 \frac{22187}{47710} = 3 + \frac{22187}{47710} \approx 3 + 0.465038 = 3.465038$.Thus,$x = 3 \frac{22187}{47710}$.

If $3^{x-2}=5$ and $\log_{10} 2=0.30103, \log_{10} 3=0.4771$,then $x=$

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