CPhill

256^(x+2)=16^(3x-1)

Notice that 256 =  2^8   and 16 = 2^4

So we have

2^8^(x + 2) = 2^4^(3x -1)   and dropping the equal bases and solving for the exponents, we have

8(x + 2) = 4(3x - 1)

8x + 16 = 12x - 4    add 4 to both sides and subtract 8x from both sides

20 = 4x    divide both sides by 4

5 = x

Don't worry about it....I checked both of the points in your equation and it was all good....in fact, I kind of like your way better than the way I learned it...LOL!!!

That's an interesting technique, saseflower.....I don't think I've seen it done that way before.....very nice....!!!!

Yep....taking out a common factor is a vital skill in being able to factor...glad you saw it...!!!

Good, minee.....now we can just substitute the values in for w and e and get

42(3) + 6(4) - 29   =  

126 + 24 - 29 =

126 - 5  =

121

No prob.....

2x(x-4)+9(x-4) and x^2(x-3)+7(x-3)

Note that (x - 4) is common to both terms in the first one...so we have

(x-4)(2x + 9)

The second one is similar......

Excellent, saseflower....

Demi  originally paid 22 x .60 = 13.20

And then she left an additional 20% of this.....so the total amount she left was.....

13.20(1.2) = $15.84

We are looking for the answer to this....

2^N = 1,000,000   take the log of each side

log2^N  = log1,000,000    and by a property of logs, we have

N log2  = log1,000,000       and log of 1,000,000 = 6    so we have

N log 2  = 6                       divide both sides by log 2

N = 6/log2 = about 20 hrs