How do you solve #4^x=13#?

1 Answer
sente
Dec 30, 2016

#x = ln(13)/ln(4) = log_4(13) ~~ 1.850#

Explanation:

Using the property of logarithms that #log(a^x) = xlog(a)#, we have

#4^x = 13#

#=> ln(4^x) = ln(13)#

#=> xln(4) = ln(13)#

#:. x = ln(13)/ln(4) = log_4(13) ~~ 1.850#

(The last line uses the fact that #log_a(b) = log(b)/log(a)#)

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