Answer:
11.776
Step-by -step explanation:
Here we are given a equation which is ,
[tex]\rm\longrightarrow 10^{x-10}-8=51.7 [/tex]
And we would like to solve this using logarithms , so firstly add 8 on both sides , we get ;
[tex]\rm\longrightarrow 10^{x-10}= 51.7+8 [/tex]
Simplify,
[tex]\rm\longrightarrow 10^{x-10}= 59.7 [/tex]
Now take log to the base 10 on both sides ,
[tex]\rm\longrightarrow log_{10}10^{x-10}=log_{10}(59.7) [/tex]
Recall that [tex]log\ m^n = n\ log m [/tex] .So ;
[tex]\rm\longrightarrow (x-10)(log_{10}10=log_{10}(59.7) [/tex]
Again recall that [tex] log_a a = 1[/tex] .So ; here [tex]log_{10}10[/tex] becomes 1 .
[tex]\rm\longrightarrow (x -10)1 = log_{10}59.7 \\ [/tex]
[tex]\rm\longrightarrow (x-10)= log_{10}(59.7)[/tex]
And the value of log 59.7 = 1.7759 . So on substituting this value , we have ;
[tex]\rm\longrightarrow x -10 = 1.7759 [/tex]
Add 10 to both sides ;
[tex]\rm\longrightarrow x = 10+1.7759 [/tex]
Simplify,
[tex]\rm\longrightarrow \underline{\underline{\red{\rm { x = 11.7759 \approx 11.776}}}}[/tex]
And we are done !
[tex]\rule{200}4[/tex]
[tex]\large\red{\bigstar}\underline{\underline{\textsf{\textbf{ Some related formulae :- }}}}[/tex]
[tex]\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\ ^{a} log \ a= 1\\\\\circ \ ^{a}log \ 1 = 0 \\\\\circ \ ^{a ^{n}} log \ b^{m}= \dfrac{m}{n} \times\:^{a}log \ b \\\\\circ \ ^{a^{m}} \ log \ b^{m} = \ ^{a}log \ b \\\\\circ \ ^{a}log \ b = \dfrac{1}{^{b}log \ a} \\\\\circ \ ^{a}log \ b = \dfrac{^{m}log \ b}{^{m} log \ a} \\\\\circ \ a^{^{a} logb} = b \\\\\circ \ ^{a}log \ b + ^{a}log \ c = \ ^{a}log(bc) \\\\\circ \ ^{a}log \ b -\: ^{a}log \ c = \ ^{a}log \left( \dfrac{b}{c} \right) \\\circ \ ^{a}log \ b \:\cdot\: ^{a}log \ c = \ ^{a}log \ c \\\\\circ \ ^{a}log \left( \dfrac{b}{c} \right) = \ ^{a}log \left(\dfrac{c}{b}\right)\end{minipage}}} [/tex]
