Question

Solve the exponential equation. Express your solutions in exact form only. Please show all of your work.
4^x-10*2^x-74=1

Answer 1

\[(2^{x})^{2}-10\times2^{x}-78=0\]
let \[2^x=t\],t>0
then \[t^2-10t-78=0\]
then solve this equation and leave out the negative value
t=\[5+\sqrt{103}\]
so x=\[\log_{2} (5+\sqrt{103})\]

Answer 2

4^x-10*2^x-74=1

4^x-10*2^x-75=0
(2)^2x -10(2)^x -75=0
(2^x)^2 -10(2^x) -75=0
m=2^x so that
m^2 -10m -75=0, factoring or solved by quadratic formula
(m+5)(m-15)=0, m=-5 and m = 15 now using m=2^x
for m=15, 2^x=15 taking log on both sides
log2^x=log15
xlog2=log15
x=log15/ log2 ans.......

if using m=-5=2^x taking log on both sides
log(-5)=log2^x therefore we cant use -5

Answer 3

which one of you is correct?