(2^-x)sinx = 1/2x - 1
My teacher said to use logs, but I'm useless at them
I would really appreaciate any help given
Thanks in advance!
Having two different transcendental functions makes that a really difficult equation, particularly with that (1/2)x- 1. Unless you mean $\displaystyle (1/2)^x$= 2^{-x}[/tex]. If so, it is still difficult but you might be able to write sin(x) as $\displaystyle \frac{e^{ix}- e^{-ix}}{2i}$ and write 2^{-x} as $\displaystyle e^{-xln(2)}$