Question Find the solution to the inequality $\log_{10}x > \log_{10}(x-1)$ .

Studdy Solution

STEP 1

Assumptions
1. We are dealing with logarithms to the base 10.
2. The inequality is $\log_{10} x > \log_{10}(x - 1)$ .
3. We assume that $x$ is in the domain of the logarithm function, which means $x > 0$ and $x - 1 > 0$ .

STEP 2

To solve the inequality $\log_{10} x > \log_{10}(x - 1)$ , we can use the property that if $\log_{a} b > \log_{a} c$ , then $b > c$ , provided that $a > 1$ , $b > 0$ , and $c > 0$ .

STEP 3

Apply the property to the given inequality.
$x > x - 1$

STEP 4

Simplify the inequality by subtracting $x$ from both sides.
$x - x > x - 1 - x$

STEP 5

This simplifies to:
$0 > -1$

STEP 6

Since $0 > -1$ is always true, the original inequality $\log_{10} x > \log_{10}(x - 1)$ holds true as long as the arguments of both logarithms are positive.

STEP 7

Recall the domain restrictions from STEP_1: $x > 0$ and $x - 1 > 0$ .

STEP 8

The second domain restriction, $x - 1 > 0$ , simplifies to $x > 1$ .

STEP 9

Combine the domain restrictions: since $x > 1$ implies $x > 0$ , the solution to the inequality is all $x$ values greater than 1.
The solution to the inequality $\log_{10} x > \log_{10}(x - 1)$ is $x > 1$ .
Therefore, the correct answer is E. $x > 1$ .

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