[tex]\\ \sf\Rrightarrow 4^x-4^{x-1}=24[/tex]
[tex]\\ \sf\Rrightarrow 4^x-4^x4^{-1}=24[/tex]
[tex]\\ \sf\Rrightarrow 4^x(1-0.25)=24[/tex]
[tex]\\ \sf\Rrightarrow 4^x(0.75)=24[/tex]
[tex]\\ \sf\Rrightarrow 4^x=32[/tex]
[tex]\\ \sf\Rrightarrow 2^{2x}=2^5[/tex]
[tex]\\ \sf\Rrightarrow 2x=5[/tex]
[tex]\\ \sf\Rrightarrow (2x)^x=(5)^x[/tex]
Answer:
Step-by-step explanation:
Let [tex]4^{x}=y[/tex], then we obtain that:
[tex]4^{x}-4^{x-1}=24\rightarrow y-(y/4)=24 \rightarrow \frac{3y}{4}=24[/tex]
So, we can get: [tex]3y=96 \rightarrow y = 96/3 = 32[/tex]
Now, because [tex]y=4^{x}=32[/tex], we can obtain the value of x as follows:
[tex]4^{x}=(2^{2})^{x}=(2^{5}) \rightarrow 2x = 5 \rightarrow x=5/2[/tex]
Then [tex](2x)^{x}=(2(5/2))^{5/2}=5^{5/2}=\sqrt{5^{5}}=25\sqrt 5[/tex]
There is no answer in your question.
