How do you solve #10^(1-4m)=18#?

Question

1484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-14T13:35:36

#color(blue)(m=(1-log_(10)(18))/4~~-0.6381812628)#

Using logarithms to the base #10#.

From the laws of logarithms:

#log_b(a^c)=clog_b(a)#

And, if:

#y=log_ba<=>b^y=a#

#log_b(b^y)=log_b(a)#

#ylog_b(b)=log_b(a)#

#y=(log_b(a))/(log_b(b)#

But: #y=log_b(a)#

#log_b(a)=(log_b(a))/(log_b(b)#

Rearranging:

#log_b(b)=(log_b(a))/(log_b(a))=1#

Using these ideas:

#10^(1-4m)=18#

Taking base 10 logarithms:

#log_(10)(10^(1-4m))=log_(10)(18)#

#(1-4m)log_(10)(10)=log_(10)(18)#

#1-4m=log_(10)(18)#

#color(blue)(m=(1-log_(10)(18))/4~~-0.6381812628)#