exponential equation with different bases; no logarithms

Question

Last week I wrote a test on exponential equations and I came across this problem;

$$10*4^x-21*10^x=10*25^x$$

But I could not found a way to solve it, because it has different bases, which I cannot make the same. Here is what I came up with:

$$10*2^{2x}-21*2^x*5^x=10*5^{2x}$$

Here I only have 2 bases. I tried to solve it with two changes:

$$2^x=a$$ $$5^x=b$$

And then I got this equation $$10a^2-21ab=10b^2$$

That is the furthest I managed to come. Of course I do not know if this is even correct.
Also I have seen on the internet such methods as using logarithms, but we have not taken this course yet, so it is not a solution I am looking for.

Answer 1

write your equation in the form $$10-21\left(\frac{5}{2}\right)^x=10\left(\left(\frac{5}{2}\right)^x\right)^2$$ and Substitute $$t=\left(\frac{5}{2}\right)^x$$ and solve a quadratic equation

Answer 2

There's nothing wrong with what you've done so far. Just to carry it on, we have $$10a^2-21ab-10b^2=0,$$ that is $$10a^2+4ab-25ab-10b^2=0,$$

that is

$$2a(5a+2b)-5b(5a+2b)=0,$$

so $$10a^2-21ab-10b^2=0 \Rightarrow (2a-5b)(5a+2b)=0.$$

This then implies

$$2a=5b,\quad 5a=-2b,$$ undoing your change of variables for the first expression

$$2\cdot 2^x=5\cdot 5^x,$$ that is $$\frac{5}{2}=\Big(\frac{2}{5}\Big)^x,$$

so $$x=-1.$$ For the second expression, we undo the change of variables $$5\cdot 2^x=-2\cdot 5^x,$$ that is $$-\frac{2}{5}=\Big(\frac{2}{5}\Big)^x,$$

which can never hold true. Thus, the solution is $$x=-1.$$