Solve 2\times(7)^(x+2)-2=0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2 \times { 7 }^{ x+2 } -2=0$$

Solve for x

$x=-2$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$2\times 7^{x+2}-2=0$$

Add $2$ to both sides of the equation.

$$2\times 7^{x+2}=2$$

Divide both sides by $2$.

$$7^{x+2}=1$$

Take the logarithm of both sides of the equation.

$$\log(7^{x+2})=\log(1)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$\left(x+2\right)\log(7)=\log(1)$$

Divide both sides by $\log(7)$.

$$x+2=\frac{\log(1)}{\log(7)}$$

By the change-of-base formula $\frac{\log(a)}{\log(b)}=\log_{b}\left(a\right)$.

$$x+2=\log_{7}\left(1\right)$$

Subtract $2$ from both sides of the equation.

$$x=-2$$

Solve for x (complex solution)

$x=\frac{2\pi n_{1}i}{\ln(7)}-2$
$n_{1}\in \mathrm{Z}$