Exponential Equations

Euler

Okay, the answer is x = 2. How did you come up with that value for x? What are the steps? So, from your answer I sense that an understanding of logs is needed to solve this equation, right?
 
Re: Euler

greatwhiteshark said:
Okay, the answer is x = 2. How did you come up with that value for x? What are the steps? So, from your answer I sense that an understanding of logs is needed to solve this equation, right?
Click to expand...

I solved it just by looking at it. I figured the answer couldn't be anything higher than 2, because 4^3 = 64, and 2^3 = 8, and 64-8-12 is not 0. If it was 1, then 4^1 = 4 - 12 would be negative. 2 works then. 4^2 = 16, - 2^2 = 4 16-4-12 = 0
 
Hello, Janet!

Can I solve the question below WITHOUT logs? If so, how?

4<sup>x</sup> - 2<sup>x</sup> - 12 = 0
.
Click to expand...
Note that: . 4<sup>x</sup> .= .(2<sup>2</sup>)<sup>x</sup> .= .(2<sup>x</sup>)<sup>2</sup>

The equation becomes: . (2<sup>x</sup>)<sup>2</sup> - 2<sup>x</sup> - 12 .= .0

Look familiar?
 
hello greatwhiteshark

4^x+2^x-12=0

take y=2^x


so we have as soroban said

(2^x)^2+2^x-12

y^2-y-12=0

solve

(y-4)(Y+3)

y=4

y=-3

2^x=4

log2^x=log4

x=log4/log2
 
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