Math Doubts

Evaluate $\dfrac{\log_{4}{17}}{\log_{9}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

In this logarithmic problem, a logarithmic expression is given and it contains two terms but they are connected by a minus symbol. Each term in this logarithmic expression is in rational form.

$\dfrac{\log_{4}{17}}{\log_{9}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

Think for an idea of simplifying expression

Compare the expressions in the numerators of the two terms and also compare the expressions in the denominators of the two terms. It givens us an idea for simplifying the given logarithmic expression.

The $\log_{4}{(17)}$ and $\log_{2}{(17)}$ are the expressions in the numerators of the terms. They are in same form and their bases are different but $\log_{4}{(17)}$ can be transformed into $\log_{2}{(17)}$ because the base of expression $\log_{4}{(17)}$ can be expressed in the form of base of expression $\log_{2}{(17)}$. Similarly, the base of $\log_{9}{(23)}$ can also be expressed in the form of the base of $\log_{3}{(23)}$ by the exponentiation.

$=\,\,\,$ $\dfrac{\log_{2^2}{17}}{\log_{3^2}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

Separate the exponents of Base of Logarithms

In the first term of the logarithmic expression, the bases of the logarithmic expressions in the numerator and denominator are in exponential notation. Hence, the exponent at base position of each logarithmic expression in the first logarithmic term can be separated by the base power rule of logarithms.

$=\,\,\,$ $\dfrac{\dfrac{1}{2} \times \log_{2}{17}}{\dfrac{1}{2} \times \log_{3}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

Simplify the Logarithmic expression

Now, we have to simplify the logarithmic expression for evaluating its value mathematically.

$=\,\,\, \require{cancel}$ $\dfrac{\cancel{\dfrac{1}{2}} \times \log_{2}{17}}{\cancel{\dfrac{1}{2}} \times \log_{3}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

$=\,\,\,$ $\dfrac{1 \times \log_{2}{17}}{1 \times \log_{3}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

$=\,\,\,$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$ $-$ $\dfrac{\log_{2}{17}}{\log_{3}{23}}$

The two terms in the above logarithmic expression are same. So, the value of first term should be equal to the value of second term. Hence, they both get cancelled each other mathematically.

$=\,\,\,$ $\cancel{\dfrac{\log_{2}{17}}{\log_{3}{23}}}$ $-$ $\cancel{\dfrac{\log_{2}{17}}{\log_{3}{23}}}$

$=\,\,\, 0$

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