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23 May, 11:41

How do I solve 4^x=19?

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  1. 23 May, 12:54
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    We can solve this using logarithms. Logarithms are written in the form [tex] / log_b (n) = p [/tex], where b is the base, n is the number, and p is the power. In the problem, 4 is the base raise to the power of x. We can take the logarithm of both sides to solve for x. [tex] / log_4 (4^{x}) = log_4 (19) [/tex] The first half of this equation can be written as: [tex] 4^{?} = 4^{x} [/tex]. The logarithm cancels out the 4, and we are left with: [tex] x = / log_4 (19) [/tex] Note: The default base for logarithms is base 10, so to get a numerical answer, type log (19) / log (4) into a calculator. This is the change of base formula. You will get 2.12396.
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