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Ok, the question is:

Solve the following for x:

$2^x4^{x-1}=70$

I have just asked Wolfram Alpha, of course though, it supplies an answer without revealing its working.

I started to try the product of the bases and the sum of the exponents, giving me:

$8^{2x-1}$ ?

But Alpha says I'm barking up the wrong tree?

With thanks in advance!

Georgina Davenport
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    $2^x 4^{x-1} = 2^{x} 2^{2x-2} = 2^{3x-2}$. – Jack D'Aurizio Oct 27 '15 at 00:53
  • Ok, I'm sorry but I'm very slow on these.

    I see you've rewritten 4 as $2^2$ I'm not clear on how the x-1 became x-2 though? I'm not even sure how you go to the next rearrangement without guessing?

     – Georgina Davenport Oct 27 '15 at 00:56
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    Remember these rules: $(n^a)^b=n^{ab}$ and $n^a n^b=n^{a+b}$ if $n$ is a positive real number.The first is used to derive $4^{x-1}=(2^2)^{x-1}=2^{2(x-1)}=2^{2x-2}$, while the second shows that $2^x 2^{2x-2}=2^{x+(2x-2)}=2^{3x-2}$. – Corellian Oct 27 '15 at 01:13
  • Ultimately you will be using logarithms. What kind of logarithms do you routinely use in your course? "Natural" logarithms, that is, base $e$, often called $ln$? Base $10$ logarithms? – André Nicolas Oct 27 '15 at 01:23
  • Andre, yep, I'll be using them. We use both bases but I'm already quite free in being able to use logs in any base to convert to logs in any other base. It's quite easy for me. – Georgina Davenport Oct 27 '15 at 01:26

3 Answers3

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The issue is that you cannot add exponents with different bases. You cannot multiply bases with different exponents. The trick is this: $4=2^2$. Then we can rewrite this as $$2^x4^{x-1}=2^x(2^2)^{(x-1)}=2^x2^{2(x-1)}=2^x2^{2x-2}.$$ Now we can add exponents: $$2^{3x-2}=70.$$ Try this.

Plutoro
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  • Coool..... thanks..... I'll get logging with it and see how I do! Thank you! I'll get back here after some experimentation. Please don't let that hold back anybody else who may have something helpful to say, thanks! – Georgina Davenport Oct 27 '15 at 00:58
  • BOOOOOMMM!!!!! Thank you soooooooo much!!!!!!!!

    Since $2^{3x-2}=70$

    I guessed that:

    $3x-2 = \log(70)/\log(2)$

    I then tried to solved for x in the usual way:

    Tested by plugging the answer back into the question and I got 70!

    I take it my working above is right?

    You lot are awesome! I thank you very deeply!!! I mean it!

     – Georgina Davenport Oct 27 '15 at 01:22
  • Ok.... there is one thing.... I haven't properly thought through Alex Ss suggestion. I get $2^x(2^2)^{(x-1)}$ But the two steps following.... I don't think I understand those power manipulations? How does $2^{2^{x-1}}$ become $2^{2(x-1)}$ Please? – Georgina Davenport Oct 27 '15 at 01:33
  • @user2067256 If $n$ is a positive real number, then $(n^a)^b=n^{(ab)}$. See that $4^{x-1}=(2^2)^{x-1}=2^{[2(x-1)]}$. This corresponds to the general formula where $n=2$, $a=2$, and $b=x-1$. – Corellian Oct 27 '15 at 01:46
  • @user2067256 You're welcome. Easily the most enthusiastic thanks I've received on Math.SE. Look's like you have the right idea. – Plutoro Oct 27 '15 at 01:48
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You are probably expected to notice that $4=2^2$ and write your expression as $2^{3x-2}$. However, this is not necessary in this case.

Let $\log$ be the logarithm to any base. Recall that if $a$ and $b$ are positive then $\log(ab)=\log a+\log b$, and that $\log(a^k)=k\log a$.

So from $2^x4^{x-1}=70$ we obtain, by taking the logs of both sides, $$x\log 2+(x-1)\log 4=\log(70).$$ It follows that $$x\log 2+x\log 4=\log(70)+\log 4,$$ and therefore $$x=\frac{\log{70}+\log 4}{\log 2+\log 4}.$$ One can get a simpler expression by noting that $4=2^2$ and therefore $\log 4=2\log 2$.

André Nicolas
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  • I like this answer too, but I can't see the steps. The first line, you've logged all the terms.... but then... ? :-) – Georgina Davenport Oct 27 '15 at 03:04
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    I left out a couple of minor algebra steps, maybe that's what you are missing. Note that $(x-1)\log 4=x\log 4-\log 4$, so we get $x\log 2+x\log 4-\log 4=\log(70)$. Bringing the $\log 4$ to the other side, or more properly adding $\log 4$ to both sides, we get $x\log 2+x\log 4=\log 70+\log 4$. Now the left side is $x(\log 2+\log 4)$, so we divide both sides by $\log 2+\log 4$ to "isolate" $x$. – André Nicolas Oct 27 '15 at 03:09
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eqn = 2^x 4^(x - 1) == 70;

For real solution using Mathematica

soln = Solve[eqn, x, Reals][[1]]

(*  {x -> (3*Log[2] + Log[5] + Log[7])/
       (3*Log[2])}  *)

Verifying that soln satisfies eqn

eqn /. soln

(*  True  *)

Wolfram|Alpha has a button for step-by-step solution. Using Wolfram|Alpha from within Mathematica

WolframAlpha["solve 2^x 4^(x\[Minus]1)==70 for x", {{"RealSolution", 2}, 
  "Content"}, PodStates -> {"RealSolution__Step-by-step solution"}]

enter image description here

This is equivalent to earlier result

(x /. soln) == 2/3 + Log[70]/(3 Log[2])

(*  True  *)
Bob Hanlon
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