How do you solve #8^(2x) = 8^(x+7)#?

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Answer 1 · 2016-04-05T03:48:04

#x=7#

Note that since the bases of both exponential functions are equal, their exponents must also be equal.

This gives us, once we set their exponents equal to one another:

#2x=x+7" "=>" "x=7#

If we want to formalize this, take the logarithm with base #8# of both sides to undo the exponential functions.

#log_8(8^(2x))=log_8(8^(x+7))" "=>" "2x=x+7#

Which is the equation we saw previously.

We can generalize this idea of exponential functions having the same bases as saying that if:

#a^b=a^c" "=>" "b=c#