# Solve the exponential equations

I know what are exponential equations, and knows how to solve some forms of exponential equations, but the equation that follows not know how to solve. Please help me. The quations is:$$4^x+9^x=2\cdot 6^x$$Thanky very much, for your help.

Divide the both side for $2^{2x}$, and earn $$1+\left(\frac{3}{2}\right)^{2x}=2\cdot\left(\frac{3}{2}\right)^{x}.$$ We now substitutes: $$\left(\frac{3}{2}\right)^{x}=t$$ we have: $$1+t^2=2t\Rightarrow t^2-2t+1=0$$ whose solutions are $$t_1=t_2=1.$$ Now go to replacing the above: $$\left(\frac{3}{2}\right)^{x}=1\Rightarrow x=0$$ Equation that you have to solve is exponential equation who transformed the quadratic equation. I hope you help.

• Quite an intuitive method of solving. – Simply Beautiful Art Apr 1 '16 at 21:22

Let $a=2^x$, $b=3^x$.

$$4^x+9^x=2\cdot6^x$$ $$(2^x)^2+(3^x)^2=2\cdot(2^x)(3^x)$$ $$a^2+b^2=2ab\iff(a-b)^2=0\iff a=b$$ $$\implies2^x=3^x\implies(\frac 23)^x=1\implies x=0$$

Hint we can create perfect square so we get $$\left(1-\left(\frac{3}{2}\right)^x\right)^2=0$$ thus $1=(1.5)^x$ now taking log we get $0=x\log(1.5)$ thus $x=0$

Let $$f(x) = 4^x + 9^x - 2\times6^x = \left(2^x - 3^x \right)^2$$

If $x \not= 0$, $f(x) \not= 0$. If $x = 0$, $f(0) = 0$. Thus $f(x)$ has exactly one root i.e. $x = 0$

This is because $2^x < 3^x$ when $x > 0$ and $2^x > 3^x$ when $x < 0$