# precalculus

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x= 3^log_3_^8 ? The log being in the exponent is throwing me off.

• precalculus -

I don't get the question-what are the dashes for?

• precalculus -

No reason, basically the equation is supposed to read 3^log base 3^8?

• precalculus -

recall the definition of logarithm:

b^log_b(n) = n

3^log_3(8) = 8

log_3(8) is the exponent of 3 which produces 8

log_10(100) = 2 because 10^2 = 100

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