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Question Number 193368 by MATHEMATICSAM last updated on 11/Jun/23
If log_a y = (1/3) and log_8 a = x + 1 then show  that y = 2^(x + 1)
$$\mathrm{If}\:\mathrm{log}_{{a}} {y}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:\mathrm{log}_{\mathrm{8}} {a}\:=\:{x}\:+\:\mathrm{1}\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{y}\:=\:\mathrm{2}^{{x}\:+\:\mathrm{1}} \\ $$
Answered by aba last updated on 11/Jun/23
log_8 a=x+1 ⇒ ln(a)=3(x+1)ln(2) ⇒ ln(a)=3ln(2^(x+1) ) (1)  log_a y=(1/3) ⇒ ln(a)=3ln(y) (2)  (1)=(2) ⇒ln(y)=ln(2^(x+1) ) ⇒ y=2^(x+1)  ✓
$$\mathrm{log}_{\mathrm{8}} \mathrm{a}=\mathrm{x}+\mathrm{1}\:\Rightarrow\:\mathrm{ln}\left(\mathrm{a}\right)=\mathrm{3}\left(\mathrm{x}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{ln}\left(\mathrm{a}\right)=\mathrm{3ln}\left(\mathrm{2}^{\mathrm{x}+\mathrm{1}} \right)\:\left(\mathrm{1}\right) \\ $$$$\mathrm{log}_{\mathrm{a}} \mathrm{y}=\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\:\mathrm{ln}\left(\mathrm{a}\right)=\mathrm{3ln}\left(\mathrm{y}\right)\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)=\left(\mathrm{2}\right)\:\Rightarrow\mathrm{ln}\left(\mathrm{y}\right)=\mathrm{ln}\left(\mathrm{2}^{\mathrm{x}+\mathrm{1}} \right)\:\Rightarrow\:\mathrm{y}=\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:\checkmark \\ $$
Answered by aba last updated on 11/Jun/23
log_a (y)=(1/3)=((log_2 (a))/(3log_2 (a))) ⇒ log_a (y)=((log_8 (a))/(log_2 (a)))  ⇒log_a (y)=((x+1)/(log_2 (a))) ⇒ln(y)=(x+1)ln(2)  ⇒y=2^(x+1)  ✓
$$\mathrm{log}_{\mathrm{a}} \left(\mathrm{y}\right)=\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{a}\right)}{\mathrm{3log}_{\mathrm{2}} \left(\mathrm{a}\right)}\:\Rightarrow\:\mathrm{log}_{\mathrm{a}} \left(\mathrm{y}\right)=\frac{\mathrm{log}_{\mathrm{8}} \left(\mathrm{a}\right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{a}\right)} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{a}} \left(\mathrm{y}\right)=\frac{\mathrm{x}+\mathrm{1}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{a}\right)}\:\Rightarrow\mathrm{ln}\left(\mathrm{y}\right)=\left(\mathrm{x}+\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\Rightarrow\mathrm{y}=\mathrm{2}^{\mathrm{x}+\mathrm{1}} \:\checkmark \\ $$

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