How do you solve #20^(9-4n)=5#?

Question

2082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T07:12:35

#n=(9-log_20 5)/4#

We know that :

#x^y=z iff y=log_xz#

so :

#20^(9-4n)=5 iff 9-4n=log_20 5 iff 4n=9-log_20 5 iff #

#n=(9-log_20 5)/4#

Answer 2 · 2017-07-20T07:55:09

#n =9/4-1/4(log5/log20)#

#n = 2.116#

In a question like this you first have to decide whether you will solve it as an exponential equation, or use logs.

#5# is certainly not a power of #20#, so logs are indicated.

#20^(9-4n) =5" "larr# find log of both sides

#log20^(color(blue)(9-4n)) =log5" "larr# apply the power law

#color(blue)((9-4n))log 20 = log5" "larr# isolate the factor with #n#

#9-4n = log5/log20" "larr# re-arrange to get #4n# positive

#9-log5/log20 = 4n" "larr# divide by 4

#1/4(9-log5/log20) = n#

#n = 9/4 - 1/4(log5/log20)#

#n = 2.116# (to 3 d.p.)