Question

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Answer 1

Answer:

log(a.b)

Step-by-step explanation:

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Answer 2

$\begin{aligned}&a^{1-2x}\cdot b^{1+2x}=a^{4+x}\cdot b^{4-x}\\ \\ \Longrightarrow\ \ &\dfrac{a^{4+x}}{a^{1-2x}}=\dfrac{b^{1+2x}}{b^{4-x}}\\ \\ \Longrightarrow\ \ &a^{3x+3}=b^{3x-3}\\ \\ \Longrightarrow\ \ &a^{3x}\cdot a^3=b^{3x}\cdot b^{-3}\\ \\ \Longrightarrow\ \ &(ab)^{\frac{1}{x}}=\dfrac{b}{a}\\ \\ \Longrightarrow\ \ &\dfrac{\log(\frac{b}{a})}{\log(ab)}=\dfrac{1}{x}\\ \\ \Longrightarrow\ \ &x\log\left(\dfrac{b}{a}\right)=\log(ab)\end{aligned}$