Question

Solve the following equations.
$7^{x+2}-\frac{1}{7}\cdot 7^{x+1}-{14.7}^{x-1}+{2.7}^x=48$

Answer 1

Considering the question to be: [Here ($\cdot$) means multiplication]

$7^{x+2}-\frac{1}{7}\cdot 7^{x+1}-14\cdot 7^{x-1}+2\cdot 7^x=48$

$\Rightarrow 7^{x+2}-(7^{-1}\cdot 7^{x+1})-(2\cdot 7\cdot 7^{x-1})+2\cdot 7^x=48$

$[\text{Using the identity}{a}^{-1}=\frac{1}{a}]$

$\Rightarrow 7^{x+2}-7^{(-1+x+1)}-(2\cdot 7^{(1+x-1)})+2\cdot 7^x=48$

$[\text{Using the identity}{a}^m\times a^n=a^{m+n}]$

$\Rightarrow 7^{x+2}-7^x-(2\cdot 7^x)+(2\cdot 7^x)=48$

$\Rightarrow 7^x\cdot 7^2-7^x=48[\text{Using the identity}{a}^{m+n}=a^m\times a^n]$

$\Rightarrow 7^x(7^2-1)=48[\text{Taking}$7^x$\text{as common}]$

$\Rightarrow 7^x\left(48\right)=48$

$\Rightarrow 7^x=1$

$\Rightarrow 7^x=7^0[\because a^0=1]$

$\therefore x=0$

$\left[\text{if base is same on both sides of equal to sign, then power (exponent) of both sides will be equal}\right]$