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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 $(2x)^x$?"

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  • $\begingroup$ You write fancy by placing maths stuff between $ signs and using functions from maths.tcd.ie/~dwilkins/LaTeXPrimer $\endgroup$
    – Felipe Jacob
    Commented Feb 15, 2014 at 3:02
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    $\begingroup$ I'd suggest retitling your question for clarity. $\endgroup$
    – Maroon
    Commented Feb 15, 2014 at 3:05
  • $\begingroup$ Do you already know what a logarithm is? $\endgroup$
    – Maximilian M.
    Commented Feb 15, 2014 at 3:06
  • $\begingroup$ 1)sorry for the bad grammar $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 3:11
  • $\begingroup$ 2)sorry, im a newb, im getting better at using the website though. $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 3:12

3 Answers 3

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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'.

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  • $\begingroup$ The reason i'm asking is because I've missed 1 1/2 weeks of school so i need some help. these answers are a bit to vague to help me, could i have a bit more of an explanation? $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 3:14
  • $\begingroup$ @user128689: Okay, but keep in mind that you probably learned to factor more than two weeks ago :). Remember, $4^x$ and $4^{x-1}$ are numbers. For example, the first one is $4$ multiplied by itself $x$ times. That statement tells you that $4^x$ is the product of other numbers. Which is smaller: $4^x$ and $4^{x-1}$? Will the smaller number divide into the larger? $\endgroup$
    – dmk
    Commented Feb 15, 2014 at 3:20
  • $\begingroup$ ooooooh, now I'm seeing where this is headed, thanks very much =D $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 3:23
  • $\begingroup$ @user128689: Let us know what answer you get. (By the way, technically, logarithms do come up, but if you haven't studied them yet, you probably won't even notice where. That won't stop you from finding the solution.) $\endgroup$
    – dmk
    Commented Feb 15, 2014 at 3:25
  • $\begingroup$ Alright, i will post when i get the Correct answer, but don't hold your breath. =D $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 3:30
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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 $x$.

We can now compute the value of $(2x)^x$ which is, since $x=\frac{5}{2}$, the value of $5^{\frac{5}{2}}$ that is to say $5^{2+\frac{1}{2}}=5^2 5^{\frac{1}{2}}=25 \sqrt{5}$

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$$4^x - 4^{x-1}=24$$ $$(2^x)^2 - \dfrac{1}{4}4^x=24$$ $$2^{2x} - \dfrac{1}{4}2^{2x}=24$$ $$4\cdot 2^{2x} - 2^{2x} = 96$$ $$4\cdot 2^{2x} - 2^{2x} - 96 = 0$$ 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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    $\begingroup$ Careful, the original question used the difference of $4^x$ and $4^{x-1}$, not the sum. $\endgroup$
    – dmk
    Commented Feb 15, 2014 at 4:46
  • $\begingroup$ I fixed the problem $\endgroup$
    – Anonymous Computer
    Commented Feb 15, 2014 at 5:21
  • $\begingroup$ Thank you! i couldn't get it but now i do, thanks again. $\endgroup$
    – Diamond Louis XIV
    Commented Feb 15, 2014 at 8:47
  • $\begingroup$ @user128689 Don't thank me by posting in the comments, but by accepting answers. Read the Pay it forward section at math.stackexchange.com/help/how-to-answer :) $\endgroup$
    – Anonymous Computer
    Commented Feb 15, 2014 at 9:59
  • $\begingroup$ $(-1)^2 = (-1)^4 $ but $2\neq 4$. $\endgroup$
    – R R
    Commented Feb 16, 2014 at 4:12

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