Solve for x natural log of 4x-1=36

$\ln (4 x - 1) = 36$

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

$e^{\ln (4 x - 1)} = e^{36}$

Rewrite $\ln (4 x - 1) = 36$ 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 $b^{y} = x$ .

$e^{36} = 4 x - 1$

Solve for $x$ .

Rewrite the equation as $4 x - 1 = e^{36}$ .

$4 x - 1 = e^{36}$

Add $1$ to both sides of the equation.

$4 x = e^{36} + 1$

Divide each term in $4 x = e^{36} + 1$ by $4$ and simplify.

Divide each term in $4 x = e^{36} + 1$ by $4$ .

$\frac{4 x}{4} = \frac{e^{36}}{4} + \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{e^{36}}{4} + \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{e^{36}}{4} + \frac{1}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{e^{36}}{4} + \frac{1}{4}$

Decimal Form:

$x = 1.07780788 \cdot 10^{15}$

Solve for x natural log of 4x-1=36

Related Questions

Support LCMGCF.com by disabling your adblocker.