#log_3(3^(x-8))=2-x# solve the equation?

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Answer 1 · 2018-06-07T14:42:44

#x = 5#

Given: #log_3(3^(x-8))=2-x#

Using the property #log_b(b^u) = u# we change the left side:

#x-8=2-x#

Solve for x:

#2x = 10#

#x = 5#

Answer 2 · 2018-06-07T20:04:58

#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!