Question

If we are an expression as \[{{4}^{x}}-{{4}^{x-1}}=24\], then \[{{\left( 2x \right)}^{x}}\] equals to?

Answer 1

Hint: At first find the value of x and then substitute it. At first, consider the equation and take \[{{4}^{x-1}}\] common, then divide it by 3. After that write 4 as \[{{2}^{2}}\] and 8 as \[{{2}^{3}}\]. Then apply the principle that if bases are the same exponents should be equal to find the value of x.

Complete step-by-step solution:
In the question we are given an equation \[{{4}^{x}}-{{4}^{x-1}}=24\] and we have to find the value of \[{{\left( 2x \right)}^{x}}\].
Before finding the value of \[{{\left( 2x \right)}^{x}}\] we will first find the value of x.
Now as we are given that,
\[{{4}^{x}}-{{4}^{x-1}}=24\]
So to proceed we have to take \[{{4}^{x-1}}\] common so we can write as,
\[{{4}^{x-1+1}}-{{4}^{x-1}}=24\]
Or, \[{{4}^{x-1}}\left( 4-1 \right)=24\]
So, the equation can be written as,
\[{{4}^{x-1}}.3=24\]
Now we will divide by 3 to both sides.
So, we get,
\[{{4}^{x-1}}=\dfrac{24}{3}\]
Or, \[{{4}^{x-1}}=8\]
Now as we know that 4 = \[{{2}^{2}}\] and 8 = \[{{2}^{3}}\]. So, we can also write \[{{4}^{x-1}}\] as \[{{2}^{2\left( x-1 \right)}}\] or \[{{2}^{2x-2}}\].
Hence, the equation can be written as,
\[{{2}^{2x-2}}={{2}^{3}}\]
Now as we know that if bases are the same exponents will be equal, so here bases are 2 which is the same on both sides, and here exponents are (2x - 2) and 3 which according to rule should be equal.
Hence, we can write,
$2x – 2 = 3$
So, $2x = 5$
Hence, the value of \[x=\dfrac{5}{2}\].
Now as we know x so we will substitute in \[{{\left( 2x \right)}^{x}}\] to get the answer which is \[{{\left( 2\times \dfrac{5}{2} \right)}^{\dfrac{5}{2}}}\] which is \[{{\left( 5 \right)}^{\dfrac{5}{2}}}\] or \[{{\left( 5 \right)}^{2}}\times {{\left( 5 \right)}^{\dfrac{1}{2}}}=25\times \sqrt{5}\] or \[25\sqrt{5}\].
Hence the value is \[25\sqrt{5}\].

Note: After simplifying the equation to \[{{2}^{2x-2}}={{2}^{3}}\], we compared the bases and equalized the exponents so instead of this method we can take logarithms to both the sides and then proceed. Also, students after finding out value x miss that they have to find the value of what is asked so be careful about it.