How do you solve # 3.14159^x=4#?

1 Answer
sente · mason m
Jul 21, 2016

#x=ln(4)/ln(3.14159)=log_3.14159(4)#

Explanation:

We will use the property of logarithms that #ln(a^x) = xln(a)#

With that:

#3.14159^x = 4#

#=> ln(3.14159^x)=ln(4)#

#=> xln(3.14159)=ln(4)#

#:. x=ln(4)/ln(3.14159)#

Note that this is eqivalent to the base #3.14159# log of #4#, a result we could have also found by taking the base #3.14159# log of both sides and applying #log_a(a^x)=x#

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