Answer $9^x = 4^x + 6^x$ to a 10th grader, who knows math until the equation of a straight line (just before calculus) [closed]

Find $$x$$: $$9^x = 4^x + 6^x$$

This was in my exam today, and I have no idea, would really help if someone taught me, I would love to know how to write math on this website and is there a way to like search a topic for a 10th grader where i can find many questions and solve them?

• The general question of solving for $x$ in $a^x+b^x=c^x$ where $a,b,c$ are just arbitrary numbers like $3^x+5^x=7^x$ or $11^x+11^x=12^x$ or other examples... this general question is not solvable using "simple" or "elementary" methods. As a tenth grader you are not expected to be able to solve random problems like this. Even experts would need to rely on things like the Lambert-W function (a non-elementary function which is particularly ugly and difficult to use in practice) or numerical methods. Sep 19 at 12:56
• @JMoravitz Actually, this case is transformable into a quadratic. Sep 19 at 12:59
• If you were given a problem like this, it was either a mistake, or they expected you to be able to find it by inspection or some special property specific to these particular numbers. We can tell right away that there should be an answer since both functions are monotonic increasing and at $x=1$ you have $4^x+6^x=4^1+6^1=10$ while $9^x=9^1=9$ is less than that, while later at $x=2$ you have $4^2+6^2=16+36=52$ while $9^2=81$ is more than that, so the answer should be somewhere between $1$ and $2$, but that the answer can't be a whole number since left would be odd while right is even. Sep 19 at 12:59
• @JMoravitz Because $6$ is the geometric mean of $4$ and $9$, this particular case does have a solution for example if you let $y=\left(\frac32\right)^x$. Familiarity with logarithms might help express this solution Sep 19 at 12:59
• @JMoravitz Thank you sir, just know that I appreciate you and aspire to become like you(very good in maths and physics) one day Sep 19 at 13:02

$$9^x = 4^x + 6^x$$

Dividing by $$4^x$$ , we get :

$$\left(\frac{9}{4}\right)^x = 1 + \left(\frac{3}{2}\right)^x$$

$$\left(\frac{3}{2}\right)^{2x} = 1 + \left(\frac{3}{2}\right)^{x}$$

Substituting $$\left(\frac{3}{2}\right)^x = t$$ , you get a quadratic equation in $$t$$, $$t^2 = 1 + t$$

Now, solve for $$t$$ and then solve for $$x$$.

• +1. How did you know, or forebode, to divide by $4^x$? I failed to think up this key step! Sep 20 at 6:42
• Because √9/4 = 6/4 Sep 20 at 7:26
• @memeguy Even if you don't divide , you'll get to that eventually. Let $X=2^x$ and $Y=3^x$. Then the equation is $Y^2=XY+X^2$ and now solve one variable for another by quadratic formula. Sep 20 at 12:31