Question

Prove the logarithmic expression as $7\log \dfrac{16}{15}+5\log \dfrac{25}{24}+3\log \dfrac{81}{80}=\log 2$ $$$$

Answer 1

We use the logarithmic identity involving quotient ${{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n$ and proceed from left hand side of the statement. We then prime factorize the composite numbers, use the logarithmic identity involving power $m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$ and logarithmic identity involving product ${{\log }_{b}}mn={{\log }_{b}}m+{{\log }_{b}}n$ to simplify until we get the express in terms of $\log 2,\log 3,\log 5$. We add the like terms to arrive at the right hand side.

Complete step-by-step solution
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $x$ is the exponent to which another fixed number, the base $b$ must be raised, to produce that number $x$, which means if ${{b}^{y}}=x$ then the logarithm denoted as log and calculated as
${{\log }_{b}}x=y$
Here $x,y,b$ are real numbers and $x$ is called the argument of the logarithm. We know that
${{\log }_{b}}b=1$
We know the logarithmic identity involving power $m\ne 0$ where $m$ is real number as
$m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}$
We also know the logarithmic identity involving product as
${{\log }_{b}}mn={{\log }_{b}}m+{{\log }_{b}}n$
We also know the logarithmic identity involving quotient as
${{\log }_{b}}\left( \dfrac{m}{n} \right)={{\log }_{b}}m-{{\log }_{b}}n$
We are given in the question a statement to prove involving logarithms which is
$7\log \dfrac{16}{15}+5\log \dfrac{25}{24}+3\log \dfrac{81}{80}=\log 2$
We proceed from the left hand side using the logarithmic identity of quotient for $m=16,n=15$ in the first term, $m=25,n=24$for the second term and $m=81,n=80$for the third term. We have,

& 7\log \dfrac{16}{15}+5\log \dfrac{25}{24}+3\log \dfrac{81}{80} \\\ & =7\log 16-7\log 15+5\log 25-5\log 24+3\log 81-3\log 80 \\\ \end{aligned}$$ We replace the arguments of the logarithm with their prime factorization to have $$\begin{aligned} & =7\log \left( 2\times 2\times 2\times 2 \right)-7\log \left( 3\times 5 \right)+5\log \left( 5\times 5 \right)-5\log \left( 2\times 2\times 2\times 3 \right) \\\ & +3\log \left( 3\times 3\times 3\times 3 \right)-3\log \left( 2\times 2\times 2\times 2\times 5 \right) \\\ & =7\log {{2}^{4}}-7\log \left( 3\times 5 \right)+5\log {{5}^{2}}-5\log \left( {{2}^{3}}\times 3 \right)+3\log {{3}^{4}}-3\log \left( {{2}^{4}}\times 5 \right) \\\ \end{aligned}$$ We use the logarithmic identity involving power in first, third and fourth term with $m=4,2,4$ respectively. We use logarithmic identity involving product with $m=3,n=5$ in second term, $m={{2}^{3}},n=3$ in the fourth term and $m={{2}^{4}},n=3$ in the sixth term. We have, $$=7\times 4\log 2-7\left( \log 3+\log 5 \right)+5\times 2\log 5-5\left( \log {{2}^{3}}+\log 3 \right)+3\log {{3}^{4}}-3\left( \log {{2}^{4}}+\log 5 \right)$$ We use the logarithmic identity involving for $\log {{2}^{3}}$ with $m=3$ and for $\log {{2}^{4}}$ with $m=4$ to have, $$\begin{aligned} & =7\times 4\log 2-7\left( \log 3+\log 5 \right)+5\times 2\log 5-5\left( 3\log 2+\log 3 \right) \\\ & +3\times 4\log 3-3\left( 4\times \log 2+\log 5 \right) \\\ & =28\log 2-7\log 3-7\log 5+10\log 5-15\log 2-5\log 3 \\\ & +12\log 3-12\log 2-3\log 5 \\\ \end{aligned}$$ Let us collect the terms multiplied with $\log 2,\log 3,\log 5$separately and take$\log 2,\log 3,\log 5$. We proceed, $$\begin{aligned} & =28\log 2-15\log 2-12\log 2-7\log 3-5\log 3+12\log 3 \\\ & -7\log 5+10\log 5 -3\log 5 \\\ & =\left( 28-15-12 \right)\log 2+\left( -7-5+12 \right)\log 3+\left( -7+10-3 \right)\log 5 \\\ & =1\cdot \log 2+0\cdot \log 3+0\cdot \log 5 \\\ & =\log 2 \\\ \end{aligned}$$ **We have $\log 2$ at the right hand side of the statement. So it is proved that $$ 7\log \dfrac{16}{15}+5\log \dfrac{25}{24}+3\log \dfrac{81}{80}=\log 2$$** **Note:** We note that we cannot find logarithms of negative numbers and 0. The base of logarithm is also always positive excluding 1. The prime factorization of a composite number $n$ with $n$ prime factors ${{p}_{1}},{{p}_{2}},...,{{p}_{m}}$ can be written as $n={{p}_{1}}^{{{e}_{1}}}{{p}_{2}}^{{{e}_{2}}}...{{p}_{m}}^{{{e}_{m}}}$ where ${{e}_{1}},{{e}_{2}},...{{e}_{m}}$are positive integral exponents. When the base is not specified either the base is 10 (base of common logarithm) or the base is $e$, base of natural logarithm