Solve by Factoring 10^(2x-3)+3=19

$10^{2 x - 3} + 3 = 19$

Step 1

Subtract $19$ from both sides of the equation.

$10^{2 x - 3} + 3 - 19 = 0$

Step 2

Subtract $19$ from $3$ .

$10^{2 x - 3} - 16 = 0$

Step 3

Add $16$ to both sides of the equation.

$10^{2 x - 3} = 16$

Step 4

Take the base $10$ logarithm of both sides of the equation to remove the variable from the exponent.

$\log (10^{2 x - 3}) = \log (16)$

Step 5

Expand the left side.

Expand $\log (10^{2 x - 3})$ by moving $2 x - 3$ outside the logarithm.

$(2 x - 3) \log (10) = \log (16)$

Logarithm base $10$ of $10$ is $1$ .

$(2 x - 3) \cdot 1 = \log (16)$

Multiply $2 x - 3$ by $1$ .

$2 x - 3 = \log (16)$

Step 6

Add $3$ to both sides of the equation.

$2 x = \log (16) + 3$

Step 7

Divide each term in $2 x = \log (16) + 3$ by $2$ and simplify.

Divide each term in $2 x = \log (16) + 3$ by $2$ .

$\frac{2 x}{2} = \frac{\log (16)}{2} + \frac{3}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{\log (16)}{2} + \frac{3}{2}$

Divide $x$ by $1$ .

$x = \frac{\log (16)}{2} + \frac{3}{2}$

Simplify the right side.

Simplify each term.

Rewrite $\log (16)$ as $\log (2^{4})$ .

$x = \frac{\log (2^{4})}{2} + \frac{3}{2}$

Expand $\log (2^{4})$ by moving $4$ outside the logarithm.

$x = \frac{4 \log (2)}{2} + \frac{3}{2}$

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 \log (2)$ .

$x = \frac{2 (2 \log (2))}{2} + \frac{3}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$x = \frac{2 (2 \log (2))}{2 (1)} + \frac{3}{2}$

Cancel the common factor.

$x = \frac{2 (2 \log (2))}{2 \cdot 1} + \frac{3}{2}$

Rewrite the expression.

$x = \frac{2 \log (2)}{1} + \frac{3}{2}$

Divide $2 \log (2)$ by $1$ .

$x = 2 \log (2) + \frac{3}{2}$

Step 8

Simplify each term.

Simplify $2 \log (2)$ by moving $2$ inside the logarithm.

$x = \log (2^{2}) + \frac{3}{2}$

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

$x = \log (4) + \frac{3}{2}$

Solve by Factoring 10^(2x-3)+3=19

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