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$$2^{2x}-2^x-6=0$$

$$3^{2x}-5\cdot 3^x+4 = 0$$

No clue how to approach this problem..

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  • $\begingroup$ Hint: What if you write $(2^x-3)(2^x+2) = 0$. Can you proceed? $\endgroup$
    – Amzoti
    Dec 4, 2013 at 22:38
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    $\begingroup$ HINT: These are both hidden quadratic equations. For the first one try substituting $y=2^x$ and for the second try $y=3^x$ $\endgroup$
    – Mufasa
    Dec 4, 2013 at 22:40
  • $\begingroup$ second equation is : $(3^x-1)(3^x-4)=0$ $\endgroup$
    – K. Rmth
    Dec 4, 2013 at 22:44

3 Answers 3

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Hint:

In each case, you have quadratics in $b^x$:

In the first, put $\;y = 2^x$. Then you have $y^2 - y - 6 = 0$.

For the second, put $\;y = 3^x$. Then you have $y^2 - 5y +4 = 0$.

Note: Both quadratic equations factor quite nicely.

$\begin{align} (1)\quad (2^x)^2 - 2^x - 6 = 0 & \overset{y = 2^x}{\quad \implies\quad} y^2 - y - 6 = (y-3)(y+2) = 0 \\ \\ &\quad\iff y = 2^x = 3 \;\text{ or }\; y = 2^x = -2\end{align}$

We can see immediately that in both solutions, $2^x$ must necessarily be positive, since $2>0$. So we're left with $$2^x = 3 \implies \ln(2^x) = \ln 3 \iff x\ln 2 = \ln 3 \iff x = \dfrac{\ln 3}{\ln 2}\tag{1}$$

$\begin{align} (2)\quad (3^x)^2 - 5(3^x) +4 = 0 & \overset{y = 3^x}{\quad \implies\quad} y^2 - 5y +4 = (y-1)(y-4) = 0 \\ \\ &\quad\iff y = 3^x = 1 \;\text{ or }\; y = 3^x = 4\end{align}$

So we have $$3^x = 1 \implies \ln(3^x) = \ln 1 \iff x\ln 3 = \overbrace{\ln 1 }^{=\,0}= 0 \iff x = 0\tag{1}$$

And we have $$3^x = 4 \implies \ln (3^x) = \overbrace{\ln 4}^{=\ln(2^2) = 2\ln 2} \iff x\ln 3 = 2\ln 2 \iff x = \dfrac {2\ln 2}{\ln 3}\tag{2}$$

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  • $\begingroup$ I'm not sure I understand. What do you do with 2^x if you substitute y for 2x? $\endgroup$
    – Kevin Li
    Dec 4, 2013 at 22:42
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    $\begingroup$ you should let lower ranking users answer this kind of question to encourage and motivate them. You are $100k$. You should be answering harder question, and let the easy ones serve as encouragement for new users. $\endgroup$
    – ILoveMath
    Dec 4, 2013 at 22:43
  • $\begingroup$ @DonAnselmo I'm not here for the rep...I'm here mostly to help the site and to help the users who visit MSE. Have you followed Brian M. Scott? He's in excess of 200K rep, and answers MANY less challenging questions. There's not just the challenge of content to consider; there is also the pedagogical challenge involved. "Being 'good' in math, by itself, does not a good teacher make." $\endgroup$
    – amWhy
    Dec 4, 2013 at 22:44
  • $\begingroup$ That is strange because I have seen @Brian M. Scott mostly answers topology and hard combinatorics questions. I dont think he answers questions of this level of dificulty. $\endgroup$
    – ILoveMath
    Dec 4, 2013 at 22:56
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    $\begingroup$ @DonAnselmo: Brian M. Scott does answer an occasional easy question or two. I see no problem with this, as this does provide good answers, even for the easier questions. I don't mean to stifle newer users, but it is not pleasant for higher rep users to have to tiptoe around questions which might be seen as too easy. I usually try to avoid easy questions, but once in a while, an easy question can be cathartic, not to mention I think I might be able to give a good answer. $\endgroup$
    – robjohn ♦
    Dec 10, 2013 at 7:59
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a. Let $2^x = m \implies m^2 - m - 6 = 0$

Solve the quadratic equation to get: $$(m+2)(m-3) = 0 \implies m = -2, \space m =3$$

But remember $2^x = m$ so: $$\therefore 2^x = -2, \space 2^x = 3$$

$\ln$ both sides of the equation to get:

$$\ln 2^x = \ln 3 \implies x\ln 2 = \ln 3 \implies x = \frac{\ln 3}{\ln 2}$$

$$\ln 2^x = \ln - 2$$ You cannot take the $\ln $ of a negative number so that solution is not applicable.

$$\therefore x = \frac{\ln 3}{\ln 2}$$

b. Let $3^x = n \implies n^2 -5n + 4 = 0$

Solve the quadratic equation to get: $$(n-4)(n-1) =0 \implies n = 1, \space n = 4$$

But remember $3^x = n$ so: $$\therefore 3^x = 1, \space 3^x = 4$$

$\ln$ both sides of the equation to get: $$\ln 3^x = \ln 1 \implies x\ln 3 = ln 1$$ but $$\ln 1 = 0 \implies x = 0$$

$$\ln 3^x = \ln 4 \implies x\ln 3 = \ln 4 \implies x = \frac{\ln 4}{\ln 3}$$

$$\therefore x = 0, \space x = \frac{\ln 4}{\ln 3}$$

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Put $2^{x}=X$ and you have a quadratic equation $X^2-X-6=0$ with solution $X_1=3$ and $X_2=-2$ and finally from $2^{x_k}=X_k$ you have $x_k=\frac{\log X_k}{\log 2}$ for $X_k\ge 0$.

The same for the second equation with $3^{x}=X$.

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