Maths Help !!!

XXRachelXX

Hey guys maths christmas test tomorrow and I'm just going through my book but got stuck on a few questions ! Any help would be greatly appreciated !!:D

Solve for x : 3e* -7 + 2e-*=0 Btw * stands for x cos I didnt know how to type something to the power of x !

2log9x=1/2 + log9(5x + 18)

PurpleFistMixer

Hm tis a while since I did exercises like this, but...

For the first one I'd let e^x (e to the x) equal y, then rewrite the equation as;
3y - 7 + (2/y) = 0, multiply across by y and continue it as a normal quadratic, just remember when you get your values for y at the end, that's just the value for e^x, so you'll have to get x again out of that.

For the second one... guess you could bring the log term on the right hand side over to the left, so you'll have 2log(9x) - log(9(5x+18)) = 1/2, then rewrite the left hand side using the law of logs to get log((9x)^2/9(5x+18)) = 1/2, and then solve from there... Might get messy. I never was too fond of logs.

MathsManiac

I suspect that the 9 in the OP's message was the base of the logs. In that case, after bringing thr logs to one side and combining them, you have:
log[base 9](x^2/(5x+18)) = 1/2
This gives:
9^(1/2) = x^2/(5x+18)
3 = x^2/(5x+18)
3(5x+18) = x^2
etc.

PurpleFistMixer

Hm I was indeed wondering what base it was to... Part of the reason I left out the end of the solution. : p