If #17^m=6#, what is m?

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Answer 1 · 2017-06-26T16:53:57

#m = ln(6)/ln(17)#

Given #17^m=6#

Use the natural logarithm on both sides:

#ln(17^m) = ln(6)#

NOTE: You can use any base logarithm you like; I have arbitrarily chosen natural logarithms.

Use the property of logarithms #ln(a^c) = cln(a)#:

#mln(17) = ln(6)#

Divide both sides by #ln(17)#

#m = ln(6)/ln(17)#