I have a log/exponential problem I need help with:
$\displaystyle 4^x + 6( 4 ^ {-x} ) = 5
$
So far the only thing i've thought of is to change 6 and 5 into a base of 4.
$\displaystyle 6 = 4^\frac {\log_6}{log_4} $
If u can help, thanks!
I have a log/exponential problem I need help with:
$\displaystyle 4^x + 6( 4 ^ {-x} ) = 5
$
So far the only thing i've thought of is to change 6 and 5 into a base of 4.
$\displaystyle 6 = 4^\frac {\log_6}{log_4} $
If u can help, thanks!
ok, here's the trick
multiply through by $\displaystyle 4^x$, we get:
$\displaystyle (4^x)^2 + 6 = 5(4^x)$
or in other words,
$\displaystyle (4^x)^2 - 5(4^x) + 6 = 0$
now do you see what to do?
to make it a bit clearer, what if we made $\displaystyle a = 4^x$, then we have
$\displaystyle a^2 - 5a + 6 = 0$
now continue
$\displaystyle (4^x)^2 - 5(4^x) + 6 = 0$
$\displaystyle (4^x-2)(4^x-3)=0$
$\displaystyle 4^x=2$
$\displaystyle 2^{2x}=2^1$
2x=1
x=1/2
Also,
$\displaystyle 4^x=3$
$\displaystyle x \ln 4 = \ln 3$
$\displaystyle x = \frac{\ln 3}{\ln 4}=0.792$
x = 1/2, 0.792
your answers are right