How do you solve $5 {log}_{10} (x - 2) = 11$ $5log_10(x-2)=11$ ?

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Answer 1 · 2018-07-03T13:20:58

$x = 10^{\frac{11}{5}} + 2 \approx 160.4893192$ $color(blue)(x=10^(11/5)+2~~160.4893192)$

$5 {log}_{10} (x - 2) = 11$ $5log_(10)(x-2)=11$

Divide by 5:

${log}_{10} (x - 2) = \frac{11}{5}$ $log_(10)(x-2)=11/5$

Raising the base to these:

$10^{{log}_{10} (x - 2)} = 10^{\frac{11}{5}}$ $10^(log_(10)(x-2))=10^(11/5)$

$x - 2 = 10^{\frac{11}{5}}$ $x-2=10^(11/5$

$x = 10^{\frac{11}{5}} + 2 \approx 160.4893192$ $x=10^(11/5)+2~~160.4893192$

How do you solve 5log10(x−2)=115log_10(x-2)=11?