How do you solve ${log}_{7} 12 = x$ $log_7 12 = x$ ?

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How do you solve ${log}_{7} 12 = x$ $log_7 12 = x$ ?

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Answer 1 · 2015-12-29T19:53:40

From $log$ $log$ definition, we have that ${log}_{b} a = c \Leftrightarrow b^{c} = a$ $log_ba=c <=>b^c=a$

Explanation:

Thus,

${log}_{7} (12) = x \Leftrightarrow 7^{x} = 12$ $log_7(12)=x <=> 7^x=12$

Now, we can apply $log$ $log$ on both sides of the equation.

${log 7}^{x} = log 12$ $log7^x=log12$

Another $log$ $log$ rule states that ${log}_{b} a^{n} = n \cdot {log}_{b} a$ $log_ba^n=n*log_ba$ , so:

$x \cdot log 7 = log 12$ $x*log7=log12$
$x = \frac{log 12}{log 7}$ $x=log12/log7$ (both on base 10 or base $e$ $e$ or whatever base you end up finding convenient! It really does not matter!)

Using a calculator, as these are not exact expressions, you can find the approximate value of

$x ≅ 1.277$ $x~=1.277$

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How do you solve ${log}_{7} 12 = x$ $log_7 12 = x$ ?

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