Question

if
$4 ^{x} - 4 ^{x - 1} = 24$
then find the value of x

Answer 1

${4}^{x} - {4}^{x - 1} = 24 \\ let \: {4}^{x} = t \\ t - \frac{t}{4} = 24 \\ 4t - t = 96 \\ 3t = 96 \\ t = 32 \\ \\ {4}^{x} = 32 \\ {2}^{2x} = 32 \\ {2}^{2x} = {2}^{5} \\ by \: compairing \\ 2x = 5 \\ x = 2.5$
$\red{\boxed{The \:value\: of \:x \:is \:2.5}}$

Hope it will help you.

Answer 2

Hey there!

To solve this give equation : $\bf{4^{x} - 4^{x - 1} = 24}$.

Factoring out this term $\bf{4^x - 4^{x - 1}}$ :

$\bf{4^{x - 1 + 1} - 4^{x - 1} = 24}$

Apply the exponential rule that is, $\bf{a^{b + c} = a^b a^c}$.

$\bf{4^1 \times 4^{x - 1} - 4^{x - 1} = 24}$

Now, factor out the common term "4^x - 1" :

$\bf{4^{x - 1} (4^1 - 1) = 24}$

$\bf{3 \times 4x - 1 = 24}$

Divide both the sides y the value of "3" :

$\bf{\dfrac{3 \times 4^{x - 1}}{3} = \dfrac{24}{3}}$

$\bf{4^{x - 1} = 8}$

Convert "4^{x - 1}" to the base of the value "2" that is:

$\bf{4^{x - 1} = (2^2)^{x - 1}; \quad (2^2)^{x - 1} = 8}$

Now, Again convert the base value of "8" to "2^3" that is:

$\bf{(2^2)^{x - 1} = 2^3}$

Apply the rule of exponential multiplication that is;

$\bf{(a^b)^c = a^{bc}}$

Here,  $\bf{(2^2)^{x - 1} = 2^{2(x - 1)}}$

If the functional value base $\bf{a^{f(x)} = a^{g(x)}}$, then the same base supporter will get cancelled powering values come to bases that is,  $\bf{f(x) = g(x)}$.

Therefore,

$\bf{2(x - 1) = 3}$

now, solve this whole equation to obtain the final value or the required answer:

$\bf{\dfrac{2(x - 1)}{2} = \dfrac{3}{2}}$

$\bf{x - 1 =\dfrac{3}{2}}$

$\bf{x - 1 + 1 = \dfrac{3}{2} + 1}$

$\bf{x = \dfrac{1 \times 2 + 3}{2}}$

$\boxed{\bf{\underline{\therefore \quad x = \dfrac{5}{2}}}}$

Which is the required answer for these types of queries.

Hope this helps you and clears your doubts to find the variable values by some conditional placements!!!!!!