Question

Solve the following equations, giving your ans correct to 3 sig. fig. thx u.

a) 2e^(2x+1) = e^(x+1) +15e
b) 2e^2x = 5e^(x+1) - 2e^2

Answer 1

so how would you go about solving this?

Answer 2

this property of exponents might come in handy:\[\large b^{n+m}=b^n*b^m\]

Answer 3

u=e^x for the 2nd

Answer 4

first one, divide both sides by "e" to get
\[2e^{2x}=e^x+15\] then perhaps
\[2e^{2x}-e^x-15=0\] a nice quadratic equation, that you can solve by factoring

Answer 5

second one
\[2e^{2x} = 5e^{x+1} - 2e^2\]
\[2e^{2x}=5e(e^x)-2e^2\]
\[2e^{2x}-5e(e^x)+2e^2=0\] quadratic formula for this one