27^4x = 81 x (1/9)^7x
1) $\displaystyle 1/a^n= a^{-n}$
2) $\displaystyle (a^n)( a^m)= a^{m+n}$
3) $\displaystyle \left(a^m)^n= a^{mn}$
4) If $\displaystyle a^n= a^m$ then n= m.
Corrected thanks to ModusPonens
To do it without log, make the base same if the variable is in the exponent. Make the exponents same, if the variable is in the base. Then equate the variable parts of both sides.
Otherwise log is the only way.
Okay, no logs.
I astutely observe that 27, 81, and 1/9 are all integer powers of 3. I can transforme the equation:
$\displaystyle (3^{3})^{4x} = 3^{4}\cdot \left[(3^{-2})\right]^{7x}$
Using properties of exponents, I can transform the equatino further.
$\displaystyle 3^{12x} = 3^{4-14x}$
Observing that if the bases are equal, and the "=" is telling the truth, the exponents must be equal.
12x = 4-14x
Addition
26x = 4
Division
x = 4/26 = 2/13
Okay, since it was previously established that logarithms are the "only way" after that, let's try something else...
I type the whole thing into MathCad. Select and 'x' and pick "Solve". Bam!! x = 2/13. It may have used logarithms, I suppose.
Maybe a nice chart.
Yes, that looks like right around x = 2/13.
Some questions just challenge the purpose of learning or teaching? We HAVE logarithms! Why avoid them? There are MANY ways to proceed. Why state that there are only two ways?
Let's keep our minds open a bit, shall we?