Answer:
The value of y = 0
Step-by-step explanation:
Given the equation
[tex]8^3\cdot 8^{-5}\cdot 8^{\:y}=8^{-2}[/tex]
Apply exponent rule: [tex]a^b\cdot \:a^c=a^{b+c}[/tex]
so
[tex]8^3\cdot \:\:8^{-5}=8^{3-5}=8^{-2}=\frac{1}{8^2}=\frac{1}{64}[/tex]
Thus, the equation becomes
[tex]\frac{1}{64}\left(8\:^y\right)=\frac{1}{64}[/tex]
Divide both sides by 1/64
[tex]\frac{\frac{1}{64}\left(8\:^y\right)}{\frac{1}{64}}=\frac{\frac{1}{64}}{\frac{1}{64}}[/tex]
[tex]8\:^y=1[/tex]
Solve Exponent
[tex]8\:^y=1[/tex]
Take log of both sides
[tex]\:log\left(8\:^y\right)\:=log\:\left(1\right)[/tex]
[tex]y\cdot \left(log\left(8\right)\right)=log\left(1\right)[/tex]
[tex]y\:=\frac{log\left(1\right)}{\left(log\left(8\right)\right)}[/tex]
[tex]y = 0[/tex]
Therefore, the value of y = 0
