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

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Answer 1 · 2018-04-14T13:35:36

$m = \frac{1 - {log}_{10} (18)}{4} \approx - 0.6381812628$ $color(blue)(m=(1-log_(10)(18))/4~~-0.6381812628)$

Using logarithms to the base $10$ $10$ .

From the laws of logarithms:

${log}_{b} (a^{c}) = c {log}_{b} (a)$ $log_b(a^c)=clog_b(a)$

And, if:

$y = {log}_{b} a \Leftrightarrow b^{y} = a$ $y=log_ba<=>b^y=a$

${log}_{b} (b^{y}) = {log}_{b} (a)$ $log_b(b^y)=log_b(a)$

$y {log}_{b} (b) = {log}_{b} (a)$ $ylog_b(b)=log_b(a)$

$y = \frac{{log}_{b} (a)}{{log}_{b} (b)}$ $y=(log_b(a))/(log_b(b)$

But: $y = {log}_{b} (a)$ $y=log_b(a)$

${log}_{b} (a) = \frac{{log}_{b} (a)}{{log}_{b} (b)}$ $log_b(a)=(log_b(a))/(log_b(b)$

Rearranging:

${log}_{b} (b) = \frac{{log}_{b} (a)}{{log}_{b} (a)} = 1$ $log_b(b)=(log_b(a))/(log_b(a))=1$

Using these ideas:

$10^{1 - 4 m} = 18$ $10^(1-4m)=18$

Taking base 10 logarithms:

${log}_{10} (10^{1 - 4 m}) = {log}_{10} (18)$ $log_(10)(10^(1-4m))=log_(10)(18)$

$(1 - 4 m) {log}_{10} (10) = {log}_{10} (18)$ $(1-4m)log_(10)(10)=log_(10)(18)$

$1 - 4 m = {log}_{10} (18)$ $1-4m=log_(10)(18)$

$m = \frac{1 - {log}_{10} (18)}{4} \approx - 0.6381812628$ $color(blue)(m=(1-log_(10)(18))/4~~-0.6381812628)$

How do you solve 10^(1-4m)=18101−4m=1810^(1-4m)=18?