7log left( frac{16}{15} right) + 5log left( f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the property $n\log a = \log a^n$,we rewrite the expression: $\log \left( \frac{16}{15} \right)^7 + \log \left( \frac{25}{24} \right)^5 + \l

Vedclass · Accepted Answer

$\log 2$

Vedclass · Answer

(C) Using the property $n\log a = \log a^n$,we rewrite the expression: $\log \left( \frac{16}{15} ight)^7 + \log \left( \frac{25}{24} ight)^5 + \log \left( \frac{81}{80} ight)^3$ $= \log \left[ \left( \frac{2^4}{3 	imes 5} ight)^7 	imes \left( \frac{5^2}{2^3 	imes 3} ight)^5 	imes \left( \frac{3^4}{2^4 	imes 5} ight)^3 ight]$ $= \log \left[ \frac{2^{28}}{3^7 	imes 5^7} 	imes \frac{5^{10}}{2^{15} 	imes 3^5} 	imes \frac{3^{12}}{2^{12} 	imes 5^3} ight]$ $= \log \left[ \frac{2^{28} 	imes 5^{10} 	imes 3^{12}}{2^{15+12} 	imes 3^{7+5} 	imes 5^{7+3}} ight]$ $= \log \left[ \frac{2^{28} 	imes 5^{10} 	imes 3^{12}}{2^{27} 	imes 3^{12} 	imes 5^{10}} ight]$ $= \log \left( 2^{28-27} ight) = \log 2$

$7\log \left( \frac{16}{15} \right) + 5\log \left( \frac{25}{24} \right) + 3\log \left( \frac{81}{80} \right) =$

Similar Questions

If $\log _3 x+\log _3 y=2+\log _3 2$ and $\log _3(x+y)=2$,then

If $\alpha = \log_e(2+\sqrt{3})$,then $\frac{\cosh \alpha}{1-\tanh \alpha} + \frac{\sinh \alpha}{1-\coth \alpha} = $

The number $\log_{2} 7$ is

Let $(x_0, y_0)$ be the solution of the following equations: $(2x)^{\ln 2} = (3y)^{\ln 3}$ and $3^{\ln x} = 2^{\ln y}$. Then $x_0$ is

Evaluate: $\log _7(\log _7\sqrt {7\sqrt {7\sqrt 7 } }) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module