MHB Calculate Real Values of $x$ in Exponential Equation

[juantheron]

Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$

My Try:: Let $2^x = a$ and $3^x = b$ , Then

$\sqrt{a^2-a\cdot b+b^2}+\sqrt{b^2-b+1}+\sqrt{a^2-a+1} = a+b+1$

Now I am struck after that

Help required

Thanks

 

[soroban]

Hello, jacks!

You are on the right track . . .

Calculate real values of $x$ in

$\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1}$
. . . $=\; 2^x+3^x+1$

My try: .Let $2^x = a$ and $3^x = b$.

Then: $\sqrt{a^2-a\cdot b+b^2}+\sqrt{b^2-b+1}+\sqrt{a^2-a+1}$
. . . . $= a+b+1$

Now I am struck after that.

At this point, I have some suspicions . . .

\begin{array}{cccc}\sqrt{a^2-a+1} &amp;=&amp; a &amp; [1] \\<br /> \sqrt{b^2-b+1} &amp;=&amp; b &amp; [2] \\<br /> \sqrt{a^2-ab+b^2} &amp;=&amp; 1 &amp; [3] \end{array}

Square [1]: .a^2-a+1 \:=\:a^2 \quad\Rightarrow\quad a \,=\,1

Square [2]: .b^2-b+1 \:=\:b^2 \quad\Rightarrow\quad b \,=\,1

Then: .a \,=\,1 \quad\Rightarrow\quad 2^x \,=\,1 \quad\Rightarrow\quad \boxed{x \,=\,0}

. . which satisfies all the equations.

 

[juantheron]

Calculation of real values of $x$ in $\sqrt{4^x-6^x+9^x}+\sqrt{9^x-3^x+1}+\sqrt{4^x-2^x+1} = 2^x+3^x+1$

My Try:: Let $2^x = a$ and $3^x = b$ , Then

$\sqrt{a^2-a\cdot b+b^2}+\sqrt{b^2-b+1}+\sqrt{a^2-a+1} = a+b+1$

Thanks Soroban I have got it.

Now I have use the Inequality $(a-b)^2\geq 0\Rightarrow a^2+b^2\geq 2ab$

Now $3a^2+3b^2\geq 6ab\Rightarrow 4a^2+4b^2-4ab\geq a^2+b^2+2ab$

So $\displaystyle a^2-ab+b^2\geq \frac{a^2+b^2+2ab}{4}\Rightarrow \displaystyle \sqrt{a^2-ab+b^2}\geq \sqrt{\frac{(a+b)^2}{4}} = \frac{a+b}{2}$

So $\displaystyle \sqrt{a^2-ab+b^2}\geq \frac{a+b}{2}$ and equality hold when $a=b$

So $\displaystyle \sqrt{a^2-ab+b^2}+\sqrt{a^2-a+1}+\sqrt{b^2-b+1}\geq a+b+1$

and equality hold when $a = b$. So $2^x = 3^x\Rightarrow x = 0$