Question

please answer this question with solution in photo attachment,I cannot understand solution in written​

Answer 1

ANSWER:

Given:

$\:\:\:\:\bullet\:\:\:\:4^x+4^x+4^x+4^x+4^x+4^x+4^x+4^x=\frac{1}{512}$

To Find:

$\:\:\:\:\bullet\:\:\:\:\text{Value of -}\frac{3}{x}$

Solution:

$\text{We are given that}\\\\:\longrightarrow 4^x+4^x+4^x+4^x+4^x+4^x+4^x+4^x=\dfrac{1}{512}\\\\\text{We know that,}\\\\:\longrightarrow a_1+a_2+...+a_n=n(a)\\\\\text{So,}\\\\:\implies4^x+4^x+4^x+4^x+4^x+4^x+4^x+4^x=\dfrac{1}{512}\\\\:\implies8(4^x)=\dfrac{1}{512}\\\\\text{We know that,}\\\\:\longrightarrow4=2\times2=2^2\:\:\:and\:\:\:8=2^3\\\\\text{Also,}\\\\:\longrightarrow512=2\times2\times2\times2\times2\times2\times2\times2\times2=2^9\\\\\text{So,}\\\\:\implies2^3(2^{2^x})=\dfrac{1}{2^9}$

$\text{We know that,}\\\\:\longrightarrow a^{m^n}=a^{m\times n}\\\\\text{And,}\\\\:\longrightarrow\dfrac{1}{a^m}=a^{-m}\\\\\text{So,}\\\\:\implies2^3(2^{2^x})=\dfrac{1}{2^9}\\\\:\implies2^3(2^{2x})=2^{-9}\\\\\text{We know that,}\\\\:\longrightarrow a^m\times a^n=a^{m+n}\\\\\text{So,}\\\\:\implies2^3(2^{2x})=2^{-9}\\\\:\implies2^{2x+3}=2^{-9}\\\\\text{As, both the bases are same, we compare the exponents,}\\\\:\implies2x+3=-9\\\\\text{Transposing 3 to RHS,}\\\\:\implies2x=-9-3\\\\:\implies2x=-12$

$\text{Transposing 2 to RHS,}\\\\:\implies x=\dfrac{-12\!\!\!\!/^{\,\:6}}{2\!\!\!/}\\\\\underline{:\implies x=-6}\\\\\text{Now, we need to find the value of,}\\\\:\longrightarrow-\dfrac{3}{x}\\\\:\implies\dfrac{-\!\!\!\!/\:\,3\!\!\!/}{-\!\!\!\!/\,\:6\!\!\!/_{\,\:2}}\\\\\text{So,}\\\\\bf{:\implies-\dfrac{3}{x}=\dfrac{1}{2}}$

Formulae Used:

$\:\:\:\:\bullet\:\:\:\:a^{x^y}=a^{xy}\\\\\:\:\:\:\bullet\:\:\:\:a^x\times a^y=a^{x+y}\\\\\:\:\:\:\bullet\:\:\:\:\frac{1}{a^x}=a^{-x}$

Learn More:

$\begin{gathered}\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Laws of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}\end{gathered}$