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

Question

1715 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:

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

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

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

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

How do you solve 10^(1-4m)=1810^(1-4m)=18?