Solve for x e^(7x-1)=(e^5)^x

$e^{7 x - 1} = {(e^{5})}^{x}$

Step 1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$e^{7 x - 1} = e^{5 x}$

Step 2

Since the bases are the same, then two expressions are only equal if the exponents are also equal.

$7 x - 1 = 5 x$

Step 3

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $5 x$ from both sides of the equation.

$7 x - 1 - 5 x = 0$

Subtract $5 x$ from $7 x$ .

$2 x - 1 = 0$

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Step 4

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 0.5$