Question

find the value of x
$( \frac{7}{5} ) ^{2x + 1} \times ( \frac{7}{5} ) ^{5} = ( \frac{7}{5} )^{x + 2}$
the value of x ok answer ​

Answer 1

EXPLANATION:-

$\sf(\frac{7}{5})^{2x+1}\times (\frac{7}{5})^5=(\frac{7}{5})^{x+2}\\\\(\frac{7}{5})^{2x+1+5}=(\frac{7}{5})^{x+2}\\\\cancelling\;then\;we\;get:-\\2x+1+5=x+2\\\\2x+6=x+2\\\\2x-x=2-6\\\\x=-4$

So the value of x is -4

EXTRA INFORMATION:-

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

Answer 2

x=-4 is answer

thnk u

<>¡:-))