Kassandra Mccall

2022-09-04

What is the solution to the equation ${9}^{x}-{6}^{x}-2\cdot {4}^{x}=0$?

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 ${2}^{x}$:

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

And then I tried

$$\begin{array}{rl}x& ={\mathrm{log}}_{3}a\\ x& ={\mathrm{log}}_{2}b\\ {\mathrm{log}}_{3}a& ={\mathrm{log}}_{2}b\end{array}$$

But I don't know what to do after this point. Any help is appreciated.

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 ${2}^{x}$:

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

And then I tried

$$\begin{array}{rl}x& ={\mathrm{log}}_{3}a\\ x& ={\mathrm{log}}_{2}b\\ {\mathrm{log}}_{3}a& ={\mathrm{log}}_{2}b\end{array}$$

But I don't know what to do after this point. Any help is appreciated.

Bernard Scott

Beginner2022-09-05Added 9 answers

HINT:

Divide by 4^x to get

$${a}^{2}-a-2=0$$

where $a={\left({\displaystyle \frac{3}{2}}\right)}^{x}$

Can you solve for a?

Now for real $x,a>0$

Divide by 4^x to get

$${a}^{2}-a-2=0$$

where $a={\left({\displaystyle \frac{3}{2}}\right)}^{x}$

Can you solve for a?

Now for real $x,a>0$

revitprojectb7

Beginner2022-09-06Added 1 answers

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

$${a}^{2}-ab-2{b}^{2}=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}b=\frac{a}{2},\text{or}b=-a.$$

You now have a system of two equations, ${\mathrm{log}}_{3}a={\mathrm{log}}_{2}b$ 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

$${\mathrm{log}}_{3}a={\mathrm{log}}_{2}\frac{a}{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a={2}^{\frac{\mathrm{ln}3}{\mathrm{ln}3/2}}$$

Then

$$x={\mathrm{log}}_{3}a=\frac{\mathrm{ln}3}{\mathrm{ln}3/2}{\mathrm{log}}_{3}2=\frac{\mathrm{ln}2}{\mathrm{ln}3/2}=\frac{\mathrm{ln}2}{\mathrm{ln}3-\mathrm{ln}2}$$

Hint: We have that

$${a}^{2}-ab-2{b}^{2}=0\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}b=\frac{a}{2},\text{or}b=-a.$$

You now have a system of two equations, ${\mathrm{log}}_{3}a={\mathrm{log}}_{2}b$ 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

$${\mathrm{log}}_{3}a={\mathrm{log}}_{2}\frac{a}{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}a={2}^{\frac{\mathrm{ln}3}{\mathrm{ln}3/2}}$$

Then

$$x={\mathrm{log}}_{3}a=\frac{\mathrm{ln}3}{\mathrm{ln}3/2}{\mathrm{log}}_{3}2=\frac{\mathrm{ln}2}{\mathrm{ln}3/2}=\frac{\mathrm{ln}2}{\mathrm{ln}3-\mathrm{ln}2}$$

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