Solve for x log base 5 of 2x-3=2

$\log_{5} (2 x - 3) = 2$

Step 1

Rewrite $\log_{5} (2 x - 3) = 2$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$5^{2} = 2 x - 3$

Step 2

Solve for $x$ .

Rewrite the equation as $2 x - 3 = 5^{2}$ .

$2 x - 3 = 5^{2}$

Raise $5$ to the power of $2$ .

$2 x - 3 = 25$

Move all terms not containing $x$ to the right side of the equation.

Add $3$ to both sides of the equation.

$2 x = 25 + 3$

Add $25$ and $3$ .

$2 x = 28$

Divide each term in $2 x = 28$ by $2$ and simplify.

Divide each term in $2 x = 28$ by $2$ .

$\frac{2 x}{2} = \frac{28}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{28}{2}$

Divide $x$ by $1$ .

$x = \frac{28}{2}$

Simplify the right side.

Divide $28$ by $2$ .

$x = 14$

Solve for x log base 5 of 2x-3=2

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