Question

Question :-
Prove that :
$\rm \: i.)7 \: log \: \dfrac{16}{15} + 5 \: log \dfrac{25}{24} + 3 \: log \: \dfrac{81}{80} = log2$
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Answer 1

EXPLANATION.

$\sf \implies 7 \ log \dfrac{16}{15} \ + 5 \ log\dfrac{25}{24} \ + 3 \ log\dfrac{81}{80} = log(2).$

As we know that,

Formula of :

$\sf \implies log_{a}MN\ = log_{a}M + log_{a}N.$

$\sf \implies \alpha \ log_{a}N = log_{a}N^{\alpha } \ \ (\alpha \ any \ real \ no).$

Using this formula in equation, we get.

$\sf \implies 7 \ log\dfrac{16}{15} = \bigg(log\dfrac{16}{15} \bigg)^{7}$

$\sf \implies 5 \ log\dfrac{25}{24} = \bigg(log\dfrac{25}{24} \bigg)^{5}$

$\sf \implies 3 \ log\dfrac{81}{80} = \bigg(log\dfrac{81}{80} \bigg)^{3}$

$\sf \implies \bigg[ log\bigg(\dfrac{16}{15} \bigg)^{7} \ + log\bigg(\dfrac{25}{24} \bigg)^{5} \ + log\bigg(\dfrac{81}{80} \bigg)^{3} \bigg]$

$\sf \implies log \bigg[ \bigg(\dfrac{16}{15} \bigg)^{7} \times \bigg(\dfrac{25}{24} \bigg)^{5} \times \bigg(\dfrac{81}{80} \bigg)^{3} \bigg]$

$\sf \implies log \bigg[ \bigg(\dfrac{2^{4} }{3 \times 5} \bigg)^{7} \times \bigg(\dfrac{5^{2} }{2^{3} \times 3} \bigg)^{5} \times \bigg(\dfrac{3 ^{4} }{2^{4} \times 5} \bigg) \bigg] \ = log(2).$

Hence proved.

                                                                                                                       

MORE INFORMATION.

$\sf (1) = log_{a}MN = log_{a}M + log_{a}N$

$\sf (2) = log_{a} \dfrac{M}{N} = log_{a}M - log_{a}N$

$\sf (3) = log_{a}N^{\alpha } = \alpha \ log_{a}N \ \ (\alpha \ any \ real \ no)$

$\sf (4) = log_{a}^{\beta } N^{\alpha } = \dfrac{\alpha }{\beta } log_{a}N \ \ (\alpha \ne 0 , \ \beta \ne 0)$

$\sf (5) = log_{a}N = \dfrac{log_{b}N}{log_{b}a}$

$\sf (6) = log_{b}a . log_{a}b = 1 = log_{b}a = \dfrac{1}{log_{a}b}$

$\sf (7) = e^{lna^{x} } =a^{x}$

Answer 2

Step-by-step explanation:

$\rm 7 \: log\dfrac{16}{15}+5\:log\dfrac{25}{24}+3\:log\dfrac{81}{80}=log\:2$

$\rm \longmapsto log\bigg(\dfrac{16}{15}\bigg)^7+log \bigg(\dfrac{25}{24}\bigg)^5+log \bigg(\dfrac{81}{80}\bigg)^3$

$\rm\rightarrow log\bigg((\dfrac{16}{15})^7\times(\dfrac{25}{24})^5\times(\dfrac{81}{80})^3\bigg)$

$\rm\longmapsto log\bigg((\dfrac{2^4}{3\times5})^7\times(\dfrac{5^2}{2^3\times3})^5\times(\dfrac{3^4}{2^4\times5})^3\bigg)$

$\rm\rightarrow log\bigg(\dfrac{2^{28}}{3^7\times5^7}\times\dfrac{5^{10}}{2^{15}\times3^5}\times\dfrac{3^{12}}{2^{12}\times5^3}\bigg)$

$\rm\rightarrow log(2^{28-15-12}\times5^{10-7-3}\times3^{12-7-5})$

$\rightarrow \rm log(2^1\times5^0\times3^0)$

$\rightarrow \rm log(2\times 1\times1)\longmapsto log \: 2$

$\sf{\underline{\boxed{ Hence, verified} \dag}}$