How do you solve $- 10 + {log}_{3} (n + 3) = - 10$ $−10 + log_ 3 (n + 3) = −10$ ?

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How do you solve $- 10 + {log}_{3} (n + 3) = - 10$ $−10 + log_ 3 (n + 3) = −10$ ?

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Answer 1 · 2016-01-25T15:11:51

$n = - 2$ $n = -2$

Explanation:

$- 10 + {log}_{3} (n + 3) = - 10$ $-10 + log_3(n+3) = -10$

First of all, add $10$ $10$ on both sides of the equation:

$\Leftrightarrow \times {log}_{3} (n + 3) = 0$ $<=> color(white)(xx) log_3(n+3) = 0$

To "get rid" of the ${log}_{3}$ $log_3$ term, you need to exponentiate the expression to the base $3$ $3$ , since $a^{x}$ $a^x$ is the inverse function for ${log}_{a} (x)$ $log_a(x)$ and thus, both $a^{{log}_{a} (x)} = x$ $a^(log_a(x)) = x$ and ${log}_{a} (a^{x}) = x$ $log_a(a^x) = x$ hold.

$\Leftrightarrow \times 3^{{log}_{3} (n + 3)} = 3^{0}$ $<=> color(white)(xx) 3^(log_3(n+3)) = 3^0$

... don't forget that for any number $b$ $b$ you can compute $b^{0} = 1$ $b^0 = 1$

$\Leftrightarrow \times \times x n + 3 = 1$ $<=> color(white)(xxxxx) n + 3 = 1$

... subtract $3$ $3$ on both sides of the equation...

$\Leftrightarrow \times \times \times n = - 2$ $<=> color(white)(xxxxxx) n = -2$

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How do you solve $- 10 + {log}_{3} (n + 3) = - 10$ $−10 + log_ 3 (n + 3) = −10$ ?

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How do you solve −10+log3(n+3)=−10−10 + log_ 3 (n + 3) = −10?

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How do you solve $- 10 + {log}_{3} (n + 3) = - 10$ $−10 + log_ 3 (n + 3) = −10$ ?