Solve for x log base 4 of 2x+1+3=5

$\log_{4} (2 x + 1) + 3 = 5$

Move all terms not containing $\log_{4} (2 x + 1)$ to the right side of the equation.

Subtract $3$ from both sides of the equation.

$\log_{4} (2 x + 1) = 5 - 3$

Subtract $3$ from $5$ .

$\log_{4} (2 x + 1) = 2$

Rewrite $\log_{4} (2 x + 1) = 2$ 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 $b^{y} = x$ .

$4^{2} = 2 x + 1$

Solve for $x$ .

Rewrite the equation as $2 x + 1 = 4^{2}$ .

$2 x + 1 = 4^{2}$

Raise $4$ to the power of $2$ .

$2 x + 1 = 16$

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

Subtract $1$ from both sides of the equation.

$2 x = 16 - 1$

Subtract $1$ from $16$ .

$2 x = 15$

Divide each term in $2 x = 15$ by $2$ and simplify.

Divide each term in $2 x = 15$ by $2$ .

$\frac{2 x}{2} = \frac{15}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{15}{2}$

Divide $x$ by $1$ .

$x = \frac{15}{2}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{15}{2}$

Decimal Form:

$x = 7.5$

Mixed Number Form:

$x = 7 \frac{1}{2}$

Solve for x log base 4 of 2x+1+3=5

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