Question

1/ log 24 base 6 +1/log 24 base 12+ 1/log 24 base 8

Answer 1

Given:

Given:[math]\log_23=x[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math][math]=\frac{\log(2^3)+\log(3)}{\log(2^2)+\log(3)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math][math]=\frac{\log(2^3)+\log(3)}{\log(2^2)+\log(3)}[/math][math]=\frac{3\log(2)+\log(3)}{2\log(2)+\log(3)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math][math]=\frac{\log(2^3)+\log(3)}{\log(2^2)+\log(3)}[/math][math]=\frac{3\log(2)+\log(3)}{2\log(2)+\log(3)}[/math][math]=\frac{\log(2)\left(3+\frac{\log(3)}{\log (2)}\right)}{\log(2)\left(2+\frac{\log(3)}{\log (2)}\right)}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math][math]=\frac{\log(2^3)+\log(3)}{\log(2^2)+\log(3)}[/math][math]=\frac{3\log(2)+\log(3)}{2\log(2)+\log(3)}[/math][math]=\frac{\log(2)\left(3+\frac{\log(3)}{\log (2)}\right)}{\log(2)\left(2+\frac{\log(3)}{\log (2)}\right)}[/math][math]=\frac{3+\frac{\log(3)}{\log (2)}}{2+\frac{\log(3)}{\log (2)}}[/math]

Given:[math]\log_23=x[/math][math]x=\log_2(3)=\frac{\log(3)}{\log(2)}[/math][math]\therefore \log_{12}(24)=\frac{\log(24)}{\log(12)}[/math][math]=\frac{\log(2^3\times 3)}{\log(2^2\times3)}[/math][math]=\frac{\log(2^3)+\log(3)}{\log(2^2)+\log(3)}[/math][math]=\frac{3\log(2)+\log(3)}{2\log(2)+\log(3)}[/math][math]=\frac{\log(2)\left(3+\frac{\log(3)}{\log (2)}\right)}{\log(2)\left(2+\frac{\log(3)}{\log (2)}\right)}[/math][math]=\frac{3+\frac{\log(3)}{\log (2)}}{2+\frac{\log(3)}{\log (2)}}[/math][math]=\frac{3+x}{2+x}[/math]

Answer 2

above answer is correct