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How do you solve #log_b64=3#?

1 Answer

Nov 10, 2016

Logarithms can be converted to exponential form using #log_a(b) = n-> a^n = b#. So,for this problem, we have:

#b^3 = 64#

Equating exponents:

#b^3 = 4^3#

#b = 4#

Hopefully this helps!

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