$\begingroup$So, every week our teacher gives us a very difficult question worth 1 point and I can never get them right, so I would highly appreciate the person who tells and explains, in good detail, why this problem solves the way it does and each step you took to find the answer. When you answer I would also greatly appreciate your taking into account my age of 16, because of this i may not comprehend some words so if u could take a bit more effort to explain your meanings, that would be fantastic. Just so you know, the class i'm in is Advanced Algebra II.
Now for the question; $4^x-4^{x-1}=24$ what is the value of $(2x)^x$?"
$\endgroup$103 Answers
$\begingroup$Hint: There's a term that you can factor from the left-hand side of your equation. What is it? What do you get if you rewrite that side as the product of two terms?
Beyond that, as I'm sure you know, $4 = 2^2$ and $8 = 2^3$. Just sayin'.
$\endgroup$5$\begingroup$Let us start with your equation which is $$4^x + 4^{x-1}=24$$ First, factor $4^x$; this gives $$ 4^x \left(1-\frac{1}{4}\right)=\frac{3\ 4^x}{4}=24$$ So $4^x=32$ which can write $2^{2x}=2^5$. So $2x=5$ and then $x=\frac{5}{2}$. So now, we know the value of $x$.
Your second question is to compute the value of $(2x)^x$ which is, since $x=\frac{5}{2}$, the value of $5^{\frac{5}{2}}$ that is to say $5^{2+\frac{1}{2}}=5^2 5^{\frac{1}{2}}=25 \sqrt{5}$
$\endgroup$$\begingroup$$$4^x - 4^{x-1}=24$$ $$(2^x)^2 - \dfrac{1}{4}4^x=24$$ $$2^{2x} - \dfrac{1}{4}2^{2x}=24$$ $$4\cdot 2^{2x} - 2^{2x} = 96$$ $$4\cdot 2^{2x} - 2^{2x} - 96 = 0$$ Let $a=2^{2x}$. Then: $$4\cdot 2^{2x} - 2{2x} - 96$$ $$=4a - a - 96 = 0$$ $$3a - 96 = 0$$ $$3a = 96$$ $$a = \dfrac{96}{3}$$ $$a = 32$$ Replace $a$ with $2^{2x}$ again. $$2^{2x} = 32$$ Equate the exponents and solve for $x$. Remember that if you have an equation $x^m = x^n$, then it logically follows that $m = n$. Note that this is not an "if and only if" statement. Special cases like $0^2=0^3$ will pop up. $$2^{2x} = 2^5$$ $$2x = 5$$ $$x = \dfrac{5}{2}$$ Finding $(2x)^x$ is easy; just plug in the value of x. $$\left[2\left(\frac{5}{2}\right)\right]^{5/2}$$ $$=5^{5/2}$$ $$=\sqrt[2]{5^5}$$ $$=\sqrt[2]{5^4}\cdot\sqrt[2]{5^1}$$ $$=5^2\cdot\sqrt{5}$$ $$=25\sqrt{5}$$ Final Answer:
When $4^x - 4^{x-1} = 24$, then $x = \dfrac{5}{2}$. So, $(2x)^x = 25\sqrt{5}$.
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