Solve for n -10+ log base 3 of n+3=-10

$- 10 + \log_{3} (n + 3) = - 10$

Move all terms not containing $\log_{3} (n + 3)$ to the right side of the equation.

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Add $10$ to both sides of the equation.

$\log_{3} (n + 3) = - 10 + 10$

Add $- 10$ and $10$ .

$\log_{3} (n + 3) = 0$

Rewrite $\log_{3} (n + 3) = 0$ 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 $b^{y} = x$ .

$3^{0} = n + 3$

Solve for $n$

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Anything raised to $0$ is $1$ .

$1 = n + 3$

Rewrite the equation as $n + 3 = 1$ .

$n + 3 = 1$

Move all terms not containing $n$ to the right side of the equation.

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Subtract $3$ from both sides of the equation.

$n = 1 - 3$

Subtract $3$ from $1$ .

$n = - 2$

Solve for n -10+ log base 3 of n+3=-10

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