MATHS Q help ?

TrustedApple

Hi there i am studying for my repeat exam in maths and i cant work out a ANS to this Q

LOG5(X+1) + LOG5(x-3) = 1


The 5 are small

brownacid

BAsic rules of logs are here, have a read through that.

http://www.purplemath.com/modules/logrules.htm

Assuming the X in your statement is a typo and it is supposed to be (x+1) then the answer is x = +4.4641 or -2.4641.

If you're still stuck have a look through your leaving cert books if you still have them, its pretty basic stuff

eoins23456

log5((x+1)*(x-3))=LOG5(5)
(x+1)*(x-3)=5
x^2-2x-8
Then solve the quadratic for x
log5(5) =1
as in 5^1=5
and ln(a) +ln (b) =ln(a*b)

MathsManiac

brownacid wrote: »

BAsic rules of logs are here, have a read through that.

http://www.purplemath.com/modules/logrules.htm

Assuming the X in your statement is a typo and it is supposed to be (x+1) then the answer is x = +4.4641 or -2.4641.

If you're still stuck have a look through your leaving cert books if you still have them, its pretty basic stuff

But if this question is a Leaving Cert question, or being done in the context of real analysis, you would probably be expected to reject the second solution, since if you substitute it into the original equation, you end up trying to evaluate the logs of negative numbers.

Also, I think there must be a slip in your solution. The two solutions to the quadratic are 4 and -2. You can verify easily that the first is a solution of the original equation and the second is not:

x=4:
log5(5) + log5(1) = 1
1+0=1 (true)

x=-2:
log5(-1) + log5(-5) = 1
(Not a solution, since these logs don't exist in the reals.)