SOLUTION: hi pls solve for x 1. log5 x+log5 (x+4)= log5(7) 2. (9/8)log5 x = 7 3. 5log4 x = -1 4. 5^(6x-7)= 9^(10x-6) thank you sonia

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: hi pls solve for x 1. log5 x+log5 (x+4)= log5(7) 2. (9/8)log5 x = 7 3. 5log4 x = -1 4. 5^(6x-7)= 9^(10x-6) thank you sonia      Log On


   

Question 271884: hi
pls solve for x
1. log5 x+log5 (x+4)= log5(7)
2. (9/8)log5 x = 7
3. 5log4 x = -1
4. 5^(6x-7)= 9^(10x-6)
thank you
sonia
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
1. log5 x+log5 (x+4)= log5(7)
x(x+4)=7
x^2+4x-7=0
x=sqrt(11)-2
.
Ed
.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B4x%2B-7+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%284%29%5E2-4%2A1%2A-7=44.

Discriminant d=44 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-4%2B-sqrt%28+44+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%284%29%2Bsqrt%28+44+%29%29%2F2%5C1+=+1.3166247903554
x%5B2%5D+=+%28-%284%29-sqrt%28+44+%29%29%2F2%5C1+=+-5.3166247903554

Quadratic expression 1x%5E2%2B4x%2B-7 can be factored:
1x%5E2%2B4x%2B-7+=+1%28x-1.3166247903554%29%2A%28x--5.3166247903554%29
Again, the answer is: 1.3166247903554, -5.3166247903554. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B4%2Ax%2B-7+%29