How to solve (7)^(y)=21

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (7)^(y)=21 | Equatix

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$${ 7 }^{ y } =21$$

$y=\log_{7}\left(21\right)\approx 1.564575034$

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

$$7^{y}=21$$

Take the logarithm of both sides of the equation.

$$\log(7^{y})=\log(21)$$

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

$$y\log(7)=\log(21)$$

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

$$y=\frac{\log(21)}{\log(7)}$$

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

$$y=\log_{7}\left(21\right)$$

$y=\frac{2\pi n_{1}i}{\ln(7)}+\log_{7}\left(21\right)$
$n_{1}\in \mathrm{Z}$