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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 using the property $\log_b(b) = 1$ and $\log_b(b^k) = k$ . $\newline$ $\log_{(5x-5)}5$ can be simplified because the base and the number are the same, so $\log_{(5x-5)}5 = 1$ . $\newline$ $\log_{((x-1)^{2})}125$ can be simplified because $125$ is a power of $5$ , specifically $5^3$ , so $\log_{((x-1)^{2})}125 = \log_{((x-1)^{2})}(5^3) = 3\cdot\log_{((x-1)^{2})}5$ .
Apply Logarithmic Properties: Apply the property of logarithms that states $\log_b(b^k) = k$ to the second term. $\newline$ Since $(x-1)^2$ is the base and $5$ is the number, we can write $\log_{(x-1)^{2}}5 = 1$ . $\newline$ Now we have $1 + 3\cdot1 > 2$ , which simplifies to $1 + 3 > 2$ .
Combine Expressions: Step $2$ (Correction): Recognize that the base of the second logarithm is $(x-1)^2$ and the argument is $125$ , which is $5^3$ . We can use the property of logarithms that states $ext{log}_b(a^k) = k ext{log}_b(a)$ to simplify the second term. So, $ext{log}_{(x-1)^{2}}125 = ext{log}_{(x-1)^{2}}(5^3) = 3 ext{log}_{(x-1)^{2}}5$ . Now we have $1 + 3 ext{log}_{(x-1)^{2}}5 > 2$ .
Divide and Solve: Combine the logarithmic expressions. $\newline$ Since we have $1 + 3\log_{(x-1)^{2}}5 > 2$ , we can subtract $1$ from both sides to isolate the logarithmic term. $\newline$ $3\log_{(x-1)^{2}}5 > 1$ .
Convert to Exponential Form: Divide both sides by $3$ to solve for the logarithmic term. $\log_{(x-1)^{2}}5 > \frac{1}{3}.$
Solve for $x$ : Convert the logarithmic inequality to an exponential inequality. $\newline$ Since $\log_{(x-1)^{2}}5 > \frac{1}{3}$ , we can write this in exponential form as $5^{\frac{1}{3}} < (x-1)^2$ .
Finish Solving: Solve the inequality for $x$ . First, find the cube root of $5$ , which is approximately $1.71$ . So, we have $1.71 < (x-1)^2$ . Now, take the square root of both sides to solve for $x-1$ . $\sqrt{1.71} < x-1$ .
Calculate Lower Bound: Finish solving for $x$ . Add $1$ to both sides to isolate $x$ . $\sqrt{1.71} + 1 < x$ .
Calculate Lower Bound: Finish solving for $x$ . Add $1$ to both sides to isolate $x$ . $\sqrt{1.71} + 1 < x$ .Calculate the numerical value for the lower bound of $x$ . $\sqrt{1.71} + 1 \approx 2.31$ . So, $x > 2.31$ .

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