how do you solve: 4(2^2x ) = 8(2^x) − 4

g_K

4(2^2x ) = 8(2^x) − 4
what is the rule regarding this?

AnonymouslyStruggling

you use substitution and solve it like a basic quadratic equation.

let u be 2^x.
4u²=8u-4
4u²-8u+4=0
4(u²-2u+1)=0
u²-2u+1=0
(u-1)²=0
u=1
therefore, 2^x=1
so x=0

g_K

AnonymouslyStruggling
lets say this is an exponential question, would I solve it the same way?

g_K

how do you go from 2^x=1 --> x=0?

Taaaa76

g_K
any number to the power of 0 equals 1

AnonymouslyStruggling

g_K

how do you go from 2x=1 --> x=0?

as Taaaa76 said, any number (excluding 0⁰ which is undefined) to the power of 0 equals 1.

this is easily proven, considering that you should have done logarithms in year 10.

2^x=1 (original equation)
log_2{1}=x (rearrange the equation)
x=log{1}/log{2}=0 (input this into your calculator)
log{1} is always 0 so log{1}/log{2}) would become 0/log{2} would become 0.

lets say this is an exponential question, would I solve it the same way?

usually, no. the best way to solve an exponential equation is by using logarithms as i did above unless the answer is obvious, which may be due to logarithmic rules (e.g. 2^x=1) or may not (e.g. 2^x=4).

hope this helps 🙂

g_K

ohh thank you