Notes: Solve the exponent equation :81^2x = 27^(4x-1).

Sunday, August 31, 2014

The equation $\displaystyle{{81}}^{{{2}{x}}}={{27}}^{{{4}{x}-{1}}}$ has to be
solved.

First equate the base of both sides of the
equation.

$\displaystyle{{81}}^{{{2}{x}}}=$
27^(4x-1)

$\displaystyle{81}={{3}}^{{4}}$ and $\displaystyle{27}=$
3^3

This gives:

$\displaystyle{{\left({{3}}^{{4}}\right)}}^{{{2}{x}}}=$
(3^3)^(4x - 1)

$\displaystyle{{3}}^{{{8}{x}}}={{3}}^{{{12}{x}-}}$
3)

Now, the exponent terms of both the sides can be
equated to solve for x.

8x = 12x -
3

4x = 3

x =
3/4

The solution of the equation is x =
3/4.

Notes