Log Q's

timmywex

stuck on these two....

1. Solve for x,

log2 x=log 4 x+6
that is log of x to base 2 is equal to log of x+6 to base 4

2. simultaneous;

log2 x - log2 2=log2 (1-y) and log2 x + log2(x+2y)=3

Pherekydes

1. Solve for x,

log2 x=log4 (x+6)

log2 x=log2 sqrt(x+6)

x = sqrt(x+6)

x^2 = x+6

x^2-x-6 =0 and solve for x


2. simultaneous;

log2 x - log2 2=log2 (1-y) and log2 x + log2(x+2y)=3

Eqn #1: log2 (x) - log2 (2) = log2 (1-y) => x/2 = 1-y => y = 1 - (x/2)

Eqn #2: log2(x) +log2(x+2y) = 3 => x(x+2y)=2^3 => x^2 +2xy = 8

Sub for y in eqn #2 and solve for x, then back substitute for x and find y.

Hope this helped.

timmywex

wait now, is the second simultaneous not x^2 + 2xy = 8?? When you bring the log2 over it becomes 2^3 which is 8?

ah, was making a stupid multipication mistake that i couldnt notice,

thanks slow coach

Pherekydes

timmywex wrote: »

wait now, is the second simultaneous not x^2 + 2xy = 8?? When you bring the log2 over it becomes 2^3 which is 8?

Yep, that's right. My silly mistake.

timmywex

Slow coach wrote: »

Yep, that's right. My silly mistake.

no probs, will let you off, if you gada seen my mistake......

Thanks!!