If 5^(2x) + 3(5)^x = 28, then what is the value of x?

Dec 7, 2016

Let $t = {5}^{x}$.

${t}^{2} + 3 t = 28$

${t}^{2} + 3 t - 28 = 0$

$\left(t + 7\right) \left(t - 4\right) = 0$

$t = - 7 \mathmr{and} 4$

Now, since $t = {5}^{x}$:

${5}^{x} = - 7 \mathmr{and} {5}^{x} = 4$

$\ln \left({5}^{x}\right) = \ln \left(- 7\right) \mathmr{and} \ln \left({5}^{x}\right) = \ln 4$

$x = \emptyset \mathmr{and} x \ln 5 = \ln 4$

$x = \ln \frac{4}{\ln} 5$

Hopefully this helps!