Question Number 226641 by fantastic2 last updated on 08/Dec/25
$${if}\:\mathrm{log}\:_{\mathrm{8}} {a}+\mathrm{log}\:_{\mathrm{4}} {b}^{\mathrm{2}} =\mathrm{5} \\ $$$$\& \\ $$$$\mathrm{log}\:_{\mathrm{8}} ^{{b}} +\mathrm{log}\:_{\mathrm{4}} {a}^{\mathrm{2}} =\mathrm{7} \\ $$$${ab}=? \\ $$
Commented by fantastic2 last updated on 08/Dec/25
$${pls}\:{ans}.\:{this}\:{came}\:{in}\:{my}\:{math} \\ $$$${exam}\:{today}.{i}\:{got}\:\mathrm{512} \\ $$
Answered by Ghisom_ last updated on 08/Dec/25
$$\frac{\mathrm{ln}\:{a}}{\mathrm{3ln}\:\mathrm{2}}+\frac{\mathrm{ln}\:{b}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{5} \\ $$$$\frac{\mathrm{ln}\:{b}}{\mathrm{3ln}\:\mathrm{2}}+\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:\mathrm{2}}=\mathrm{5} \\ $$$$\mathrm{2}\:\mathrm{linear}\:\mathrm{equations}\:\mathrm{for}\:\begin{pmatrix}{\mathrm{ln}\:{a}}\\{\mathrm{ln}\:{b}}\end{pmatrix} \\ $$$$\Rightarrow \\ $$$$\mathrm{ln}\:{a}\:=\mathrm{6ln}\:\mathrm{2}\:\wedge\mathrm{ln}\:{b}\:=\mathrm{3ln}\:\mathrm{2} \\ $$$${a}=\mathrm{64}\wedge{b}=\mathrm{8} \\ $$$${ab}=\mathrm{512} \\ $$
Commented by fantastic2 last updated on 08/Dec/25
$${thank}\:{you} \\ $$
Answered by fantastic2 last updated on 08/Dec/25
$$\mathrm{log}\:_{\mathrm{8}} {a}+\mathrm{log}\:_{\mathrm{4}} {b}^{\mathrm{2}} =\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:_{\mathrm{2}} {a}+\frac{\mathrm{2}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{2}} {b}=\mathrm{5} \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{2}} {a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}=\mathrm{5}\Rightarrow\mathrm{32}={a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}\:..{i} \\ $$$$\mathrm{log}\:_{\mathrm{8}} ^{{b}} +\mathrm{log}\:_{\mathrm{4}} {a}^{\mathrm{2}} =\mathrm{7} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:_{\mathrm{2}} {b}+\frac{\mathrm{2}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{2}} {a}=\mathrm{7} \\ $$$$\Rightarrow\mathrm{log}\:_{\mathrm{2}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} {a}=\mathrm{7}\Rightarrow{ab}^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{128}…{ii} \\ $$$${i}×{ii} \\ $$$$\mathrm{128}×\mathrm{32}=\left({ab}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} \\ $$$$\Rightarrow{ab}=\mathrm{2}^{\mathrm{12}×\frac{\mathrm{3}}{\mathrm{4}}} =\mathrm{2}^{\mathrm{9}} =\mathrm{512} \\ $$
Commented by fantastic2 last updated on 08/Dec/25
$${i}\:{did}\:{it}\:{like}\:{this} \\ $$