Question

(4/3)^1-x = e^x (solve exponential equation)

Answer 1

we are given

$(\frac{4}{3} )^{1-x}=e^x$

Since, we have to solve for x

so, firstly we will isolate x on any one side

we can take ln on both sides

$ln((\frac{4}{3} )^{1-x})=ln(e^x)$

now, we can simplify it

$(4-x)ln((\frac{4}{3} )=x$

now, we can move all x terms altogether

$4*ln(\frac{4}{3})-xln(\frac{4}{3})=x$

$xln(\frac{4}{3})+x=4*ln(\frac{4}{3})$

$x(ln(\frac{4}{3})+1)=4ln(\frac{4}{3})$

so, we get

$x=\frac{4ln(\frac{4}{3})}{(ln(\frac{4}{3})+1)}$...........Answer