5

I want to solve: $$9^x - 6^x - 2\cdot 4^x = 0 $$

I was able to get to the equation below by substituting $a$ for $3^x$ and $b$ for $2^x$:

$$ a^2 - ab - 2b^2 = 0 $$

And then I tried

\begin{align}x &= \log_3{a} \\ x &= \log_2{b} \\ \log_3{a} &=\log_2{b}\end{align} But I don't know what to do after this point. Any help is appreciated.

Zain Patel
  • 16,802
Mertcan Ekiz
  • 209

4 Answers4

10

HINT:

Divide by $4^x$ to get $$a^2-a-2=0$$ where $a=\left(\dfrac32\right)^x$

Can you solve for $a?$

Now for real $x,a>0$

See also : Exponent Combination Laws

lab bhattacharjee
  • 274,582
  • Thanks so much. But how do I see intuitively what to divide / substitute in these cases? Does it happen over time? – Mertcan Ekiz Jul 10 '15 at 11:08
  • @MertcanEkiz, Probably yes. See also, https://math.stackexchange.com/questions/788757/find-real-solutions-for-a-mod-equation-with-power/788763#788763 – lab bhattacharjee Jul 10 '15 at 11:08
  • @MertcanEkiz: You had the "hint" when you were able to reformulate the equation which involved $x$ as the only unknown (appearing as an exponent) into the quadratic in two unknowns $a,b$. This was "homogeneous in degree", which means that the approach of dividing through by $b^2$ gets you back to an equation in one unknown, namely $a/b$ (in the notation of your Question). That's the equation that lab starts from. – hardmath Jul 10 '15 at 11:15
5

The more elegant way to solve this equation was given by lab bhattacharjee, but I'll simply elaborate on how to solve it using your method, which was still correct.

Hint: We have that $$a^2 - ab -2b^2 = 0 \iff b = \frac{a}{2}, \text{or }b= -a.$$

You now have a system of two equations, $\log_3 a = \log_2 b$ and from this you can see that neither $a$ or $b$ can be negative. So you need to rule out $b=-a$ as a solution. Instead, we take $b= a/2$. Substituting this into the logarithmic equation gives $$\log_3 a = \log_2 \frac{a}{2} \implies \large a = 2^{\frac{\ln 3}{\ln 3/2}}$$

Then $$x = \log_3 a = \frac{\ln 3}{\ln 3/2} \log_3 2 = \frac{\ln 2}{\ln 3/2} = \frac{\ln 2}{\ln 3 - \ln 2}$$

Zain Patel
  • 16,802
2

Hint: Can you factorise the quadratic?

Mark Bennet
  • 100,194
  • 2
    This is more of a comment than an answer. – hardmath Jul 10 '15 at 11:00
  • @hardmath It is a hint in response to "I don't know what to do ... Any help is appreciated." To give more substance - if you are able to reduce to a quadratic expression, factorising is always worth considering since it reduces one quadratic to two linear. In other cases completing the square is worth testing. – Mark Bennet Jul 10 '15 at 11:05
0

It is a homogeneous quadratic polynomial in two variables. The first thing to do is to de-homogenise it: $$a^2-ab-2b^2=b^2\biggl(\frac{a^2}{b^2}-\frac ab-2\biggr)$$ So setting $t=\dfrac ab$, you have to solve $\;t^2-t-2=0$, which has $-1$ as a root, hence the other is $2$, en the relation between $a$ and $b$ is: $$a=-b\quad\text{or}\quad a=2b$$ As $x$ is a real number, the first solution can't happen. So we have: $$3^x=2^{x+1}\iff 2=x\ln 3=(x+1)\ln2\iff x=\frac{\ln2}{\ln 3-\ln 2}.$$

Bernard
  • 175,478