Question

can someone solve this please​

Answer 1

Topic :-

Exponents and Powers

Given :-

$4^x-4^{x-1}-4^{x-2}=22$

To Find :-

Value of 'x'.

Formula to be Used :-

$\sf {a^{m-n}=\dfrac{a^m}{a^n}}$

Concept to be Used :-

$\sf{If\:\:a^x=a^y,\:\:then\:x=y.}$

Solution :-

$\sf {4^x-4^{x-1}-4^{x-2}=22}$

From mentioned formula, we can write it as,

$\sf {4^x-\dfrac{4^x}{4}-\dfrac{4^x}{4^2}=22}$

$\sf {4^x(1-\dfrac{1}{4}-\dfrac{1}{16})=22}$

$\sf {4^x(\dfrac{16-4-1}{16})=22}$

$\sf {4^x(\dfrac{11}{16})=22}$

$\sf {4^x=22 \times \dfrac{16}{11}}$

$\sf {4^x=32}$

$\sf {2^{2x}=2^5}$

Now,

We can say that,

2x = 5

$\sf{x=\dfrac{5}{2}}$

Answer :-

$\sf{So,\:\:x=\dfrac{5}{2}}$

Answer 2

Question:-

✯ Solve for 'x' : 4ˣ - 4ˣ⁻¹ - 4ˣ⁻² = 22

Solution:-

${ \sf➸ \: {4 {}^{x} - 4 {}^{x - 1} - 4 {}^{x - 2} = 22}}$

$\sf➸ \: {4 {}^{x} (1 - 4 {}^{ - 1} - 4 {}^{ - 2} ) = 22}$

$\sf{➸ \: 4 {}^{x} (1 - \frac{1}{4} - \frac{1}{16} ) = 22}$

${ \sf{➸ \: 4 {}^{x} ( \frac{16 - 4 - 1}{16} ) = 22}}$

${ \sf{➸ \: \:4 {}^{x} \times \frac{ \cancel{11}}{16} = \cancel22}}$

$\sf{ ➸ \: \frac{ 4 {}^{x} }{4 {}^{2} } = 2}$

$\sf➸ \: {4 {}^{x - 2} = 2}$

${ \sf➸ \: {2 {}^{2(x - 2)} = 2}}$

$\sf➸ \: {2x - 4 = 1}$

$➸ \: {\boxed{\sf {x = {\frac{5}{2} }}} \: \: {\sf{Ans.}}}$