${log}_{3} (3^{x - 8}) = 2 - x$ $log_3(3^(x-8))=2-x$ solve the equation?

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Answer 1 · 2018-06-07T14:42:44

$x = 5$ $x = 5$

Given: ${log}_{3} (3^{x - 8}) = 2 - x$ $log_3(3^(x-8))=2-x$

Using the property ${log}_{b} (b^{u}) = u$ $log_b(b^u) = u$ we change the left side:

$x - 8 = 2 - x$ $x-8=2-x$

Solve for x:

$2 x = 10$ $2x = 10$

$x = 5$ $x = 5$

Answer 2 · 2018-06-07T20:04:58

$x = 5$ $x=5$

On the left side, since we have the same base, we can rewrite our expression as the following:

$cancel(log_3)(cancel3^(x-8))=2-x$

$=>x-8=2-x$

Now this is just a simple algebra equation. Let's add $x$ to both sides to get

$2x-8=2$

Adding $8$ to both sides, we get

$2x=10$

Lastly, we divide both sides by $2$ to get

$x=5$

Hope this helps!

log_3(3^(x-8))=2-xlog3(3x−8)=2−xlog_3(3^(x-8))=2-x solve the equation?