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Question Number 25593 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

$$\mathit{i}\mathit{f}:{\mathit{l}\mathit{o}\mathit{g}}_6^{45}=\mathit{x},{\mathit{l}\mathit{o}\mathit{g}}_{18}^{24}=\mathit{y}$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{n}:\mathit{l}\mathit{o}\mathit{g}{}_{12}^{25}=?(\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{x},\mathrm{y})$$

Answered by Rasheed.Sindhi last updated on 14/Dec/17

$${\log }_{6}45=\mathrm{x}\Rightarrow 6^{\mathrm{x}}=45\Rightarrow 2^{\mathrm{x}}.3^{\mathrm{x}}=3^2.5..........\left(\mathrm{i}\right)$$$${\log }_{18}24=\mathrm{y}\Rightarrow {18}^{\mathrm{y}}=24\Rightarrow 2^{\mathrm{y}}.3^{2\mathrm{y}}=2^3.3........\left(\mathrm{ii}\right)$$$${\log }_{12}25=\mathrm{z}\left(\mathrm{say}\right)\Rightarrow {12}^{\mathrm{z}}=25\Rightarrow 2^{2\mathrm{z}}.3^{\mathrm{z}}=5^2.....\left(\mathrm{iii}\right)$$$$\mathrm{Determining}\mathrm{z}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{x},\mathrm{y}.$$$$\left(\mathrm{i}\right)\times \left(\mathrm{ii}\right)\times \left(\mathrm{iii}\right):$$$$(2^{\mathrm{x}}.3^{\mathrm{x}})(2^{\mathrm{y}}.3^{2\mathrm{y}})(2^{2\mathrm{z}}.3^{\mathrm{z}})=(3^2.5)(2^3.3)\left(5^2\right)$$$$2^{\mathrm{x}+\mathrm{y}+2\mathrm{z}}.3^{\mathrm{x}+2\mathrm{y}+\mathrm{z}}=2^3.3^3.5^3$$$$2^{\mathrm{x}+\mathrm{y}}.2^{2\mathrm{z}}.3^{\mathrm{x}+2\mathrm{y}}.3^{\mathrm{z}}=2^3.3^3.5^3$$$${\left(2^2\right)}^{\mathrm{z}}{\left(3\right)}^{\mathrm{z}}=\frac{2^3.3^3.5^3}{2^{\mathrm{x}+\mathrm{y}}.3^{\mathrm{x}+2\mathrm{y}}}$$$${12}^{\mathrm{z}}=2^{3-\mathrm{x}-\mathrm{y}}.3^{3-\mathrm{x}-2\mathrm{y}}.5^3$$$$\log \left({12}^{\mathrm{z}}\right)=\log (2^{3-\mathrm{x}-\mathrm{y}}.3^{3-\mathrm{x}-2\mathrm{y}}.5^3)$$$${\log }_{12}25=\frac{\log (2^{3-\mathrm{x}-\mathrm{y}}.3^{3-\mathrm{x}-2\mathrm{y}}.5^3)}{\log \left(12\right)}$$$$\mathrm{Or}$$$${\log }_{12}25={\log }_{12}(2^{3-\mathrm{x}-\mathrm{y}}.3^{3-\mathrm{x}-2\mathrm{y}}.5^3)$$

Commented by Rasheed.Sindhi last updated on 14/Dec/17

$$\mathrm{The}\mathrm{answer}\mathrm{has}\mathrm{been}\mathrm{corrected}.$$

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