# Additional Math -Logarithm – Change of Base and Solve for x.

Addition Math Tuition on Sunday at Woodlands Centre

You have to be familiar with the law of logarithm, especially change of base $latex {{\log }_{a}}b=\frac{{{{{\log }}_{x}}b}}{{{{{\log }}_{x}}a}}$

It is good that you also know how to ‘swap base $latex {{\log }_{a}}b=\frac{{{{{\log }}_{b}}b}}{{{{{\log }}_{b}}a}}=\frac{1}{{{{{\log }}_{b}}a}}$ The base and the number changes place with 1 as the numerator.

Please check if your values of x are accepted of rejected. Substitute the values of x back into the equation, log40 and log80 has no solution so we reject x=0 and only accept x=5/2

$latex {{\log }_{4}}({{x}^{2}}+5x)-{{\log }_{8}}{{x}^{3}}=\frac{1}{{{{{\log }}_{3}}4}}$

Step 1: Change the base of the logarithms on the left from 4 to 2

$latex \frac{{{{{\log }}_{2}}({{x}^{2}}+5x)}}{{{{{\log }}_{2}}4}}-\frac{{{{{\log }}_{2}}{{x}^{3}}}}{{{{{\log }}_{2}}8}}=\frac{1}{{{{{\log }}_{3}}4}}$

Step 2: Change the base of the logarithms on the right from 3 to 2

$latex \frac{{{{{\log }}_{2}}({{x}^{2}}+5x)}}{{2{{{\log }}_{2}}2}}-\frac{{{{{\log }}_{2}}{{x}^{3}}}}{{3{{{\log }}_{2}}2}}=\frac{{\frac{1}{{{{{\log }}_{2}}4}}}}{{{{{\log }}_{2}}3}}$

$latex \frac{1}{2}{{\log }_{2}}({{x}^{2}}+5x)-\frac{1}{3}{{\log }_{2}}{{x}^{3}}=\frac{{{{{\log }}_{2}}3}}{{{{{\log }}_{2}}4}}$

$latex \frac{1}{2}{{\log }_{2}}({{x}^{2}}+5x)-{{\log }_{2}}x=\frac{{{{{\log }}_{2}}3}}{{{{{\log }}_{2}}{{2}^{2}}}}$

Step 3: Multiply both sides by 2

$latex 2(\frac{1}{2}{{\log }_{2}}({{x}^{2}}+5x)-{{\log }_{2}}x)=2(\frac{{{{{\log }}_{2}}3}}{2})$

$latex {{\log }_{2}}({{x}^{2}}+5x)-2{{\log }_{2}}x={{\log }_{2}}3$

$latex {{\log }_{2}}\frac{{({{x}^{2}}+5x)}}{{{{x}^{2}}}}={{\log }_{2}}3$

Step 4: Remove the log on both sides

$latex {{x}^{2}}+5x=3{{x}^{2}}$

Step 5: Solve for x

$latex 0=2{{x}^{2}}-5x$

$latex x(2x-5)=0$

$latex x=0\ or\ 2x-5=0$

$latex x=0\ (reject)\ or\ x=\frac{5}{2}$

Additional Math (A-Math) ( amath) Tuition  Choa Chu Kang, Yew Tee, Yishun and Sembawang

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