Change all log to base 10 (lg). Use the change of base formula.

\displaystyle \frac{{\lg 7}}{{\lg 2}}\times \frac{{\lg 8}}{{\lg 3}}\div \frac{{\lg \sqrt{7}}}{{\lg 9}}

\displaystyle \frac{{\lg 7}}{{\lg 2}}\times \frac{{\lg {{2}^{3}}}}{{\lg 3}}\div \frac{{\lg {{7}^{{\frac{1}{2}}}}}}{{\lg {{3}^{2}}}}

Use the formula logabc= c logab. Simplify the expression.

\displaystyle \frac{{\lg 7}}{{\lg 2}}\times \frac{{3\lg 2}}{{\lg 3}}\div \frac{{\frac{1}{2}\lg 7}}{{2\lg 3}}

\displaystyle\frac{{3\lg 7}}{{\lg 3}}\div \frac{{\frac{1}{2}\lg 7}}{{2\lg 3}}

\displaystyle\frac{{3\lg 7}}{{\lg 3}}\times \frac{{2\lg 3}}{{\frac{1}{2}\lg 7}}

\displaystyle\frac{3}{1}\times \frac{2}{{\frac{1}{2}}}

12 (Shown)

 

Normally student will change the expression to the same base number. But it will be difficuil for the bove question. A better solution is to change all to base 10 (lg), the cancel out the common logs. You may also change to ln (loge) instead of lg., the answer will still be the same.

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