# Evaluate $7\log_{2}{\bigg(\dfrac{16}{15}\bigg)}$ $+$ $5\log_{2}{\bigg(\dfrac{25}{24}\bigg)}$ $+$ $3\log_{2}{\bigg(\dfrac{81}{80}\bigg)}$

The $7$ times binary logarithm of the quotient of $16$ by $15$ is added to the $5$ times the logarithm of $25$ by $24$ to base $2$. The sum of them is added to the $3$ times the binary logarithm of $81$ by $80$ in this logarithmic expression.

$7\log_{2}{\bigg(\dfrac{16}{15}\bigg)}$ $+$ $5\log_{2}{\bigg(\dfrac{25}{24}\bigg)}$ $+$ $3\log_{2}{\bigg(\dfrac{81}{80}\bigg)}$

Let us learn how to find the value of the logarithmic expression by simplification.

### Expand each term in the expression

Each term in this logarithmic expression consists of binary logarithmic expression. The quantity inside each binary log expression in each term of the expression is a rational number. Each rational number cannot be simplified further. Hence, it is recommendable to expand each term by using the quotient rule of logarithms.

$=\,\,\,$ $7\big(\log_{2}{16}-\log_{2}{15}\big)$ $+$ $5\big(\log_{2}{25}-\log_{2}{24}\big)$ $+$ $3\big(\log_{2}{81}-\log_{2}{80}\big)$

$=\,\,\,$ $7 \times \big(\log_{2}{16}-\log_{2}{15}\big)$ $+$ $5 \times \big(\log_{2}{25}-\log_{2}{24}\big)$ $+$ $3 \times \big(\log_{2}{81}-\log_{2}{80}\big)$

Each factor can be distributed over the subtraction in each term by the distributive property of multiplication across the difference of the terms.

$=\,\,\,$ $7 \times \log_{2}{16}$ $-$ $7 \times \log_{2}{15}$ $+$ $5 \times \log_{2}{25}$ $-$ $5 \times \log_{2}{24}$ $+$ $3 \times \log_{2}{81}$ $-$ $3 \times \log_{2}{80}$

$=\,\,\,$ $7\log_{2}{16}$ $-$ $7\log_{2}{15}$ $+$ $5\log_{2}{25}$ $-$ $5\log_{2}{24}$ $+$ $3\log_{2}{81}$ $-$ $3\log_{2}{80}$

### Express each quantity in exponential notation

The expansion is also not useful at this time for simplifying the binary logarithmic expression. Therefore, it is recommendable to express every quantity inside each logarithm expression in exponential form by factorization (or factorisation).

$=\,\,\,$ $7\log_{2}{(2 \times 8)}$ $-$ $7\log_{2}{(3 \times 5)}$ $+$ $5\log_{2}{(5 \times 5)}$ $-$ $5\log_{2}{(2 \times 12)}$ $+$ $3\log_{2}{(3 \times 27)}$ $-$ $3\log_{2}{(2 \times 40)}$

$=\,\,\,$ $7\log_{2}{(2 \times 2 \times 4)}$ $-$ $7\log_{2}{(3 \times 5)}$ $+$ $5\log_{2}{(5 \times 5)}$ $-$ $5\log_{2}{(2 \times 2 \times 6)}$ $+$ $3\log_{2}{(3 \times 3 \times 9)}$ $-$ $3\log_{2}{(2 \times 2 \times 20)}$

$=\,\,\,$ $7\log_{2}{(2 \times 2 \times 2 \times 2)}$ $-$ $7\log_{2}{(3 \times 5)}$ $+$ $5\log_{2}{(5 \times 5)}$ $-$ $5\log_{2}{(2 \times 2 \times 2 \times 3)}$ $+$ $3\log_{2}{(3 \times 3 \times 3 \times 3)}$ $-$ $3\log_{2}{(2 \times 2 \times 2 \times 10)}$

$=\,\,\,$ $7\log_{2}{(2 \times 2 \times 2 \times 2)}$ $-$ $7\log_{2}{(3 \times 5)}$ $+$ $5\log_{2}{(5 \times 5)}$ $-$ $5\log_{2}{(2 \times 2 \times 2 \times 3)}$ $+$ $3\log_{2}{(3 \times 3 \times 3 \times 3)}$ $-$ $3\log_{2}{(2 \times 2 \times 2 \times 2 \times 5)}$

The factoring process is completed and it is time to express the product in exponential form.

$=\,\,\,$ $7\log_{2}{(2^4)}$ $-$ $7\log_{2}{(3 \times 5)}$ $+$ $5\log_{2}{(5^2)}$ $-$ $5\log_{2}{(2^3 \times 3)}$ $+$ $3\log_{2}{(3^4)}$ $-$ $3\log_{2}{(2^4 \times 5)}$

### Evaluate the Log expression by simplification

Some quantities in log expressions are in exponential notation and they can be simplified by the power law of logarithms. Similarly, some quantities in logarithmic expressions are in product form. Hence, they can be expanded by using the product rule of logarithms.

$=\,\,\,$ $7 \times 4 \times \log_{2}{2}$ $-$ $7\big(\log_{2}{3}+\log_{2}{5}\big)$ $+$ $5 \times 2 \times \log_{2}{5}$ $-$ $5\big(\log_{2}{(2^3)}+\log_{2}{3}\big)$ $+$ $3 \times 4 \times \log_{2}{(3)}$ $-$ $3\big(\log_{2}{(2^4)}+\log_{2}{5}\big)$

$=\,\,\,$ $28 \times \log_{2}{2}$ $-$ $7 \times \log_{2}{3}$ $-$ $7 \times \log_{2}{5}$ $+$ $10 \times \log_{2}{5}$ $-$ $5 \times \log_{2}{(2^3)}$ $-$ $5 \times \log_{2}{3}$ $+$ $12 \times \log_{2}{3}$ $-$ $3 \times \log_{2}{(2^4)}$ $-$ $3 \times \log_{2}{5}$

$=\,\,\,$ $28\log_{2}{2}$ $-$ $7\log_{2}{3}$ $-$ $7\log_{2}{5}$ $+$ $10\log_{2}{5}$ $-$ $5 \times 3 \times \log_{2}{2}$ $-$ $5\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $3 \times 4 \times \log_{2}{2}$ $-$ $3\log_{2}{5}$

$=\,\,\,$ $28\log_{2}{2}$ $-$ $7\log_{2}{3}$ $-$ $7\log_{2}{5}$ $+$ $10\log_{2}{5}$ $-$ $15 \times \log_{2}{2}$ $-$ $5\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $12 \times \log_{2}{2}$ $-$ $3\log_{2}{5}$

$=\,\,\,$ $28\log_{2}{2}$ $-$ $7\log_{2}{3}$ $-$ $7\log_{2}{5}$ $+$ $10\log_{2}{5}$ $-$ $15\log_{2}{2}$ $-$ $5\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $12\log_{2}{2}$ $-$ $3\log_{2}{5}$

Now, write the logarithmic expression by writing the like logarithmic terms closer.

$=\,\,\,$ $28\log_{2}{2}$ $-$ $15\log_{2}{2}$ $-$ $12\log_{2}{2}$ $-$ $7\log_{2}{3}$ $-$ $5\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $7\log_{2}{5}$ $+$ $10\log_{2}{5}$ $-$ $3\log_{2}{5}$

It is time to focus on simplifying the binary logarithmic expression.

$=\,\,\,$ $28\log_{2}{2}$ $-$ $15\log_{2}{2}$ $-$ $12\log_{2}{2}$ $-$ $7\log_{2}{3}$ $-$ $5\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $7\log_{2}{5}$ $-$ $3\log_{2}{5}$ $+$ $10\log_{2}{5}$

$=\,\,\,$ $28\log_{2}{2}$ $-$ $27\log_{2}{2}$ $-$ $12\log_{2}{3}$ $+$ $12\log_{2}{3}$ $-$ $10\log_{2}{5}$ $+$ $10\log_{2}{5}$

$=\,\,\,$ $\log_{2}{2}$ $-$ $\cancel{12\log_{2}{3}}$ $+$ $\cancel{12\log_{2}{3}}$ $-$ $\cancel{10\log_{2}{5}}$ $+$ $\cancel{10\log_{2}{5}}$

$=\,\,\,$ $\log_{2}{2}$

The logarithm of $2$ to base to $2$ is equal to one as per the base logarithmic rule.

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

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