Question

please solve question 3

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

Let' solve the given equation,

${4}^{x} - {4}^{x - 1} = 24 \\ \\ {4}^{x} - {4}^{x} \times {4}^{ - 1} = 24 \\ \\ {4}^{x} (1 - {4}^{ - 1} ) = 24 \\ \\ {4}^{x} (1 - \frac{1}{4} ) = 24 \\ \\ {4}^{x} \times \frac{3}{4} = 24 \\ \\ {4}^{x} = 24 \times \frac{4}{3} \\ \\ {2}^{2x} = 32 \\ \\ {2}^{2x} = {2}^{5} \\ \\ 2x = 5 \\ \\ x = \frac{5}{2}$

$\textsf{\red{ Hence value of x is 5 / 2 }}$

$\textbf{ Have great future ahead! }$

Answer 2

Hey there!

We're given the expression of a value raised to a variable of "x". Here's the solution to this question, via exponential rules and cancellation of common factors.

$\bf{Expression \: \: is: \: 4^x - 4^{x - 1} = 24]$

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

Apply the following exponential rule, that is:

$\boxed{\bf{Exponential \: \: Rule: \: a^{b + c} = a^b \times a^c}}$

Here, $\bf{4^{x - 1 + 1} = 4^1 \times 4^{x - 1}}$

Now, just factor out the common term in this newly formed expression.

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

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

Divide both the current given sides by a value of "3":

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

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

Now, just do the conversion of the value "4" to a base of "2" raised to the power of "2", that is, in an exponential form. Therefore;

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

Similarly, do this with the value of "8", convert it to the base of "2" raised to a power of "3", in a exponential form.

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

Apply the exponential rule for raised multiplication that is,

$\boxed{\bf{Exponential \: \: Rule: \: \big(a^b \big)^c = a^{b \times c}}}$

$\bf{\therefore \quad Here, \: \: \: \: (2^2)^{x - 1} = 2^{2 \times (x - 1)}}$

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

This'll mean, we've equal bases, that is, according to the rule of similar bases the exponential functions are equivalent and the same base gets cancelled or equalled, that is:

$\boxed{\bf{Base \: \: Rule: \: a^{f(x)} = a^{g(x)}; \quad then, \: \: f(x) = g(x)}}$

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

Divide both the sides with a value of "2", to cancel out the multiplier in numerator.

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

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

Add the value of "1" to both the sides.

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

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

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

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

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

Which is the required solution for this type of query.

Hope this helps you and clears your doubt in heavily detailed manner for exponent applications!!!!!!!