Solving an exponential inequality $4^x + 25^x + 0.01^x \lt 2$ without using logarithms

Question

I need to solve this inequality without logarithms , 10th class

$$4^x + 25^x + 0.01^x \lt 2$$

I got $$2^{2x} + 5^{2x} + 0.1^{2x} <2$$ then $$5^{2x} + 0.1^{2x} < 2(1-2^{2x-1}),$$ so after this I realized that the left side is greater than $0$, and so $1-2^{2x-1}$ should be greater than $0$ , and by logic I got that $x$ is less than $0.5$. This is all I could do. P.S thank u all

Answer 1

By AM-GM $$4^x+25^x+\frac{1}{100^x}\geq3\sqrt[3]{4^x25^x\cdot\frac{1}{100^x}}=3>2.$$ Can you end it now?

Answer 2

The answer: no solution.

At the 10th level, all you should be expected to know is:

• $a^m>0$ for $a>0$
• $a^m$ appears to increase if $a>1$
• $a^m$ appears to decrease if $0<a<1$.

Since you are summing powers with bases greater than $1$ and powers with bases less than $1$, there is no way for the sum of three powers to be less than $2$. In general, if you are summing an $n$ number of powers that contain bases both greater and less than $1$, then the sum is always going to be greater than $n-1$.

For an extreme example, view a graph on Desmos here.

Answer 3

I think I realized how he wants you to do the problem but it is by no means obvious:

$4^x +25^x+10^{-2x} = 2^{2x}+5^{2x}+10^{-2x}= 2^{2x}+5^{2x}+5^{-2x}2^{-2x}$

Now let $u=2^{2x}$ and $v=5^{2x}$

Essentially you are solving the equation: $$ u+v +u^{-1}v^{-1} < 2 \Rightarrow u+v<2uv \Rightarrow u-2uv+v <0 \Rightarrow (u-v)^2<0$$

But this cannot be true for any real numbers u and v. Since I am assuming you are only searching for real solutions for x the same holds for $(2^{2x}-5^{2x})^2<0$

And as such there are no real solutions.