Solve for x:
log 8log 2×log 3log3=2logx\dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log}\sqrt{3}} = 2\text{log} xlog 2log 8×log3log 3=2logx
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Given,
⇒log 8log 2×log 3log3=2logx⇒log 23log 2×log 3log312=2logx⇒3log 2log 2×log 312log 3=2 log x⇒3×2=2log x⇒log x=62⇒log x=3⇒x=103=1000.\Rightarrow \dfrac{\text{log 8}}{\text{log 2}} \times \dfrac{\text{log 3}}{log\sqrt{3}} = 2log x \\[1em] \Rightarrow \dfrac{\text{log 2}^3}{\text{log 2}} \times \dfrac{\text{log 3}}{\text{log} 3^{\dfrac{1}{2}}} = 2log x \\[1em] \Rightarrow \dfrac{\text{3log 2}}{\text{log 2}} \times \dfrac{\text{log 3}}{\dfrac{1}{2}\text{log 3}} = \text{2 log x} \\[1em] \Rightarrow 3 \times 2 = 2\text{log } x \\[1em] \Rightarrow \text{log } x = \dfrac{6}{2} \\[1em] \Rightarrow \text{log } x = 3 \\[1em] \Rightarrow x = 10^3 = 1000.⇒log 2log 8×log3log 3=2logx⇒log 2log 23×log321log 3=2logx⇒log 23log 2×21log 3log 3=2 log x⇒3×2=2log x⇒log x=26⇒log x=3⇒x=103=1000.
Hence, x = 1000.
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log3x - log32 = 1
x = log 125log 25\dfrac{\text{log 125}}{\text{log 25}}log 25log 125
Given 2log10x + 1 = log10250, find
(i) x
(ii) log102x
If log xlog 5=log y2log 2=log 9log13\dfrac{\text{log x}}{\text{log 5}} = \dfrac{\text{log y}^2}{\text{log 2}} = \dfrac{\text{log 9}}{\text{log}\dfrac{1}{3}}log 5log x=log 2log y2=log31log 9, find x and y.