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log_((5x-5))5+log(x-1)^(2)125 > 2

$\log _{(5 x-5)} 5+\log (x-1)^{2} 125>2$

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Convert between exponential and logarithmic form: all bases

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Q.

\log _{(5 x-5)} 5+\log (x-1)^{2} 125>2

Simplify Logarithmic Expressions: Simplify the logarithmic expressions. $\newline$ We have the inequality $\log_{(5x-5)}5 + \log_{(x-1)^{2}}125 > 2$ . We can simplify the logarithmic expressions using the property $\log_b(b) = 1$ and $\log_b(b^k) = k$ . $\newline$ For the first term, $\log_{(5x-5)}5$ , since the base $(5x-5)$ and the argument $(5)$ are not the same, we cannot simplify this term directly. $\newline$ For the second term, $\log_{(x-1)^{2}}125$ , we can write $125$ as $5^3$ and use the property of logarithms to bring the exponent out front: $\log_{(x-1)^{2}}5^3 = 3 \cdot \log_{(x-1)^{2}}5$ . $\newline$ Since the base $\log_b(b) = 1$ $0$ and the argument $\log_b(b) = 1$ $1$ are not the same, we cannot simplify this term directly either.
Combine Logarithmic Terms: Apply the properties of logarithms to combine the terms. $\newline$ We can use the property of logarithms that states $\log_b(m) + \log_b(n) = \log_b(m*n)$ to combine the two terms into a single logarithmic expression. $\newline$ Combine the terms: $\log_{(5x-5)}5 + 3 \times \log_{(x-1)^2}5 = \log_{(5x-5)}5 + \log_{(x-1)^2}5^3 = \log_{(5x-5)}5 + \log_{(x-1)^2}125$ . $\newline$ Since the bases are different, we cannot combine these terms using the logarithm properties. Therefore, we cannot simplify the inequality further at this point.

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