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Find all real solutions to the equation:$$4^x+6^{x^2}=5^x+5^{x^2}$$

Obviously $0$ and $1$ satisfy this equation, but here $x$ is a real number, and I do not know how to do the next step, in fact, I guess this equation is not in addition to $0$ and $1$ other than the root. But you know, I need a proof! Could anyone help me? Thanks a lot!

2 Answers 2

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For any $x \in \mathbb{R}$, apply MVT to $t^{x^2}$ on $[5,6]$ and $t^x$ on $[4,5]$, we find a pair of numbers $\xi \in (5,6)$ and $\eta \in (4,5)$ (both dependent on $x$) such that

$$6^{x^2} - 5^{x^2} = x^2 \xi^{x^2-1}\quad\text{ and }\quad 5^x - 4^x = x\eta^{x-1}$$ This implies $$ {\tt LHS}-{\tt RHS} = (4^{x} + 6^{x^2}) - (5^x + 5^{x^2}) = x^2\xi^{x^2-1} - x\eta^{x-1} $$

There are 3 cases we need to study:

  • $x \in (1,\infty)$ - Both $x^2 - 1$ and $x - 1$ are positive, we have $$\begin{align} x^2\xi^{x^2-1} & \ge x^2\inf\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\ x\eta^{x-1} & \le x\sup\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1} \end{align}\\ \implies {\tt LHS} - {\tt RHS} \ge x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{> 1}\underbrace{5^{x^2-x}}_{>1} - 1)\underbrace{x5^{x-1}}_{>0} > 0$$

  • $x \in (0,1)$ - Both $x^2-1$ and $x-1$ are negative, we have

$$\begin{align} x^2\xi^{x^2-1} & \le x^2\sup\{ t^{x^2-1} : t \in (5,6) \} = x^2 \left(\inf\{ t : t \in (5,6)\}\right)^{x^2-1} = x^2 5^{x^2-1}\\ x\eta^{x-1} & \ge x\inf\{ t^{x-1} : t \in (4,5)\} = x\left(\sup\{ t : t \in (4,5)\}\right)^{x-1} = x 5^{x-1} \end{align}\\ \implies {\tt LHS} - {\tt RHS} \le x^2 5^{x^2-1} - x5^{x-1} = (\underbrace{x}_{< 1}\underbrace{5^{x^2-x}}_{<1} - 1)\underbrace{x5^{x-1}}_{>0} < 0$$

  • $x \in (-\infty,0)$ - We have $$x^2 \xi^{x^2-1} > 0 \land x \eta^{x-1} < 0 \quad\implies\quad {\tt LHS} - {\tt RHS} = x^2\xi^{x^2-1} - x\eta^{x-1} > 0$$

Combine these 3 cases, we have ${\tt LHS} \ne {\tt RHS}$ for $x \in \mathbb{R} \setminus \{ 0, 1 \}$. Since we know $0, 1$ are roots of the equation at hand, $0$ and $1$ are all the real roots of the equation.

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    A few observations that simplify your solution: 1) $5^{x^2}[(\frac{6}{5})^{x^2})-1]=5^x-4^x$ involve $x\geq0$; 2) For $x>1, 5^x-4^x=5^{x^2}[(\frac{6}{5})^{x^2})-1] >5^{x}[(\frac{6}{5})^{x})-1 =6^x-5^x$ sau $2\cdot5^x> 6^x+4^X$ false! (from MTV). It remains the case $[0, 1]$ arising from your solution. – 2017-01-02
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Let $f(x)=4^x+6^{x^2}-5^x-5^{x^2}$.

$f'(x)=0$ for $x_1=0.7655...$ only and $f$ is a decreasing function on $(-\infty,x_1]$,

$f$ is an increasing function on $[x_1,+\infty)$,

which says that our equation has two roots maximum and we are done

because $0$ and $1$ are roots.

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    Nice work!But solving the equation: $f'(x)=0$ is not easy unless you are using a computer – 2017-01-02
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    @Zuo I used a calculator. – 2017-01-02
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    @Michael Rozenberg About the same idea but without the computer: http://math.stackexchange.com/questions/1076497/no-of-different-real-values-of-x-which-satisfy-17x9x2-23x3x2/1076930#1076930 – 2017-01-03