How do you solve #3^b=17#?

2 Answers
Mar 15, 2018

Answer:

#b=2.5789#

Explanation:

Lets take the logarithm of both sides of the equation:

#b*log3=log17#

then divide both sides by #log3#:

#b=log17/log3#

My pocket calculator (HP 15C) reads

#b=2.5789#

Mar 15, 2018

Answer:

Real solution:

#b = ln 17 / ln 3#

Complex solutions:

#b = (ln 17 + 2kpi i)/ ln 3" "# for any integer #k#

Explanation:

Given:

#3^b = 17#

Note that #e^(2kpi i) = 1# for any integer #k#.

So, if #e^a = b# then #a = ln b + 2kpi i# for any integer #k#.

So while we find the real solution by taking the real valued natural log, we can also add any integer multiple of #2pi i# to find all the complex solutions too...

Take natural log of both sides of the given equation to get:

#b ln 3 = ln 17 color(grey)(+ 2kpi i)#

Divide both sides by #ln 3# to get:

#b = (ln 17 color(grey)(+2kpi i))/ln 3#