Math Doubts

Evaluate $\log{2}$ $+$ $16\log{\Big(\dfrac{16}{15}\Big)}$ $+$ $12\log{\Big(\dfrac{25}{24}\Big)}$ $+$ $7\log{\Big(\dfrac{81}{80}\Big)}$

A logarithmic expression is defined mathematically by adding four quantities in logarithmic form. We have to simplify this logarithmic expression to find the value of this expression in this logarithmic problem.

$\log{2}$ $+$ $16\log{\Big(\dfrac{16}{15}\Big)}$ $+$ $12\log{\Big(\dfrac{25}{24}\Big)}$ $+$ $7\log{\Big(\dfrac{81}{80}\Big)}$

Expand the Logarithmic terms for simplification

In the second, third and fourth terms of the expression, the quantities in ratio form inside the logarithm cannot be the successful division. In the same way, it is not possible to simplify the whole expression. Hence, it is recommendable to expand the logarithmic term in which the quantity is in ratio form and it can be done by the quotient rule of logarithms.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \Big(\log{16}-\log{15}\Big)$ $+$ $12 \times \Big(\log{25}-\log{24}\Big)$ $+$ $7 \times \Big(\log{81}-\log{80}\Big)$

The multiplication can be distributed across the subtraction by the distributive property.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \log{16}$ $-$ $16\ \times \log{15}$ $+$ $12 \times \log{25}$ $-$ $12 \times \log{24}$ $+$ $7 \times \log{81}$ $-$ $7 \times \log{80}$

Express the Quantities in exponential form

The quantity inside the first logarithmic term is a prime factor. Hence, it is recommendable to express the quantities inside the remaining logarithmic terms as the product of prime factors.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \log{\Big(2 \times 2 \times 2 \times 2\Big)}$ $-$ $16\ \times \log{\Big(3 \times 5\Big)}$ $+$ $12 \times \log{\Big(5 \times 5\Big)}$ $-$ $12 \times \log{\Big(2 \times 2 \times 2 \times 3\Big)}$ $+$ $7 \times \log{\Big(3 \times 3 \times 3 \times 3\Big)}$ $-$ $7 \times \log{\Big(2 \times 2 \times 2 \times 2 \times 5\Big)}$

Now, express the product form of the factors in exponential notation.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \log{\Big(2^4\Big)}$ $-$ $16\ \times \log{\Big(3 \times 5\Big)}$ $+$ $12 \times \log{\Big(5^2\Big)}$ $-$ $12 \times \log{\Big(2^3 \times 3\Big)}$ $+$ $7 \times \log{\Big(3^4\Big)}$ $-$ $7 \times \log{\Big(2^4 \times 5\Big)}$

Simplify the Logarithmic expression

In this logarithmic expression, some quantities are in product form inside some logarithmic terms. They can be expanded by the product rule of logarithms. It helps us in simplifying the logarithmic expression further.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \log{\Big(2^4\Big)}$ $-$ $16\ \times \Big(\log{3}+\log{5}\Big)$ $+$ $12 \times \log{\Big(5^2\Big)}$ $-$ $12 \times \Big(\log{(2^3)+\log{3}\Big)}$ $+$ $7 \times \log{\Big(3^4\Big)}$ $-$ $7 \times \Big(\log{(2^4)+\log{5}}\Big)$

Now, the multiplication can be distributed over the addition of the terms by the distributive property.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times \log{\Big(2^4\Big)}$ $-$ $16\ \times \log{3}$ $-$ $16\ \times \log{5}$ $+$ $12 \times \log{\Big(5^2\Big)}$ $-$ $12 \times \log{\Big(2^3\Big)}$ $-$ $12 \times \log{3}$ $+$ $7 \times \log{\Big(3^4\Big)}$ $-$ $7 \times \log{\Big(2^4\Big)}$ $-$ $7 \times \log{5}$

The exponential form quantities in logarithmic terms can be simplified by the power rule of logarithms.

$=\,\,\,$ $\log{2}$ $+$ $16\ \times 4 \times \log{2}$ $-$ $16\ \times \log{3}$ $-$ $16\ \times \log{5}$ $+$ $12 \times 2 \times \log{5}$ $-$ $12 \times 3 \times \log{2}$ $-$ $12 \times \log{3}$ $+$ $7 \times 4 \times \log{3}$ $-$ $7 \times 4 \times \log{2}$ $-$ $7 \times \log{5}$

$=\,\,\,$ $\log{2}$ $+$ $64 \times \log{2}$ $-$ $16\ \times \log{3}$ $-$ $16\ \times \log{5}$ $+$ $24 \times \log{5}$ $-$ $36 \times \log{2}$ $-$ $12 \times \log{3}$ $+$ $28 \times \log{3}$ $-$ $28 \times \log{2}$ $-$ $7 \times \log{5}$

It is time to write like logarithmic terms closer for simplifying the logarithmic expression.

$=\,\,\,$ $\log{2}$ $+$ $64 \times \log{2}$ $-$ $36 \times \log{2}$ $-$ $28 \times \log{2}$ $-$ $16\ \times \log{3}$ $-$ $12 \times \log{3}$ $+$ $28 \times \log{3}$ $-$ $16\ \times \log{5}$ $+$ $24 \times \log{5}$ $-$ $7 \times \log{5}$

We can now take the common factors out from the terms and it helps us to evaluate the simplified logarithmic expression.

$=\,\,\,$ $\log{2} \times (1+64-36-28)$ $+$ $\log{3} \times (-16-12+28)$ $+$ $\log{5} \times (-16+24-7)$

$=\,\,\,$ $(1+64-36-28) \times \log{2}$ $+$ $(-16-12+28) \times \log{3}$ $+$ $(-16+24-7) \times \log{5}$

$=\,\,\,$ $(1) \times \log{2}$ $+$ $(0) \times \log{3}$ $+$ $(1) \times \log{5}$

$=\,\,\,$ $1 \times \log{2}$ $+$ $0 \times \log{3}$ $+$ $1 \times \log{5}$

$=\,\,\,$ $\log{2}$ $+$ $\log{5}$

The sum of the logarithmic terms can be added by the product rule of logarithms.

$=\,\,\,$ $\log{\Big(2 \times 5\Big)}$

$=\,\,\,$ $\log{10}$

According to the logarithmic mathematics, the common logarithm of ten is one.

$=\,\,\,$ $1$

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