Question

Can someone please help me with this, thank you

Answer 1

Topic :-

Logarithm

Given :-

$\sf{a=\left(\dfrac{1}{9}\right)^{-2\log_37}}$

$\sf {b=2^{-\log_{\frac{1}{2}}7}}$

$\sf{a=b^k}$

To Find :-

Value of 'k'.

Solution :-

$\sf{a=\left(\dfrac{1}{9}\right)^{-2\log_37}}$

$\sf{a=\left(\dfrac{1}{3^2}\right)^{-2\log_37}}$

$\sf{a=(3^{-2})^{-2\log_37}}$

$\sf {\left(\because \dfrac{1}{a^m}=a^{-m}\right)}$

$\sf{a=3^{(-2(-2))\log_37}}$

$\sf {\left(\because (a^m)^n=a^{mn}\right)}$

$\sf{a=3^{4\log_37}}$

$\sf{a=7^{4\log_33}}$

$\sf {\left(\because m^{r\log n}=n^{r\log m}\right)}$

$\boxed{\sf{a=7^4}}$

$\sf {(\because \log_mm=1)}$

$\sf {b=2^{-\log_{\frac{1}{2}}7}}$

$\sf {b=2^{-\log_{2^{-1}}7}}$

$\sf {b=2^{\frac{-1}{-1}\log_{2}7}}$

$\sf {\left(\because \log_{p^q}r=\dfrac{1}{q}\log_{p}r\right)}$

$\sf {b=2^{\log_{2}7}}$

$\sf {b=7^{\log_{2}2}}$

$\sf {\left(\because m^{\log n}=n^{\log m}\right)}$

$\sf {b=7^1}$

$\boxed{\sf{b=7}}$

$\sf {(\because \log_mm=1)}$

Now,

$\sf{a=b^k}$

Substituting values,

$\sf{7^4=7^k}$

Equating powers,

k = 4

Answer :-

Value of k = 4, hence option (A) is correct option.