Solve for y natural log of 2y-1=3

$\ln (2 y - 1) = 3$

To solve for $y$ , rewrite the equation using properties of logarithms.

$e^{\ln (2 y - 1)} = e^{3}$

Rewrite $\ln (2 y - 1) = 3$ 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 $b^{y} = x$ .

$e^{3} = 2 y - 1$

Solve for $y$ .

Rewrite the equation as $2 y - 1 = e^{3}$ .

$2 y - 1 = e^{3}$

Add $1$ to both sides of the equation.

$2 y = e^{3} + 1$

Divide each term in $2 y = e^{3} + 1$ by $2$ and simplify.

Divide each term in $2 y = e^{3} + 1$ by $2$ .

$\frac{2 y}{2} = \frac{e^{3}}{2} + \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{e^{3}}{2} + \frac{1}{2}$

Divide $y$ by $1$ .

$y = \frac{e^{3}}{2} + \frac{1}{2}$

The result can be shown in multiple forms.

Exact Form:

$y = \frac{e^{3}}{2} + \frac{1}{2}$

Decimal Form:

$y = 10.54276846 \dots$

Solve for y natural log of 2y-1=3

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