SOLUTION: 1) Express loga X2 + 2 loga Square root of X as a single logarithm. Simplify if possible. 2) Given logb 2=0.387 and logb 5=0.898 find log square root of 20 3) solve 2x+3

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 1) Express loga X2 + 2 loga Square root of X as a single logarithm. Simplify if possible. 2) Given logb 2=0.387 and logb 5=0.898 find log square root of 20 3) solve 2x+3       Log On


   



Question 38201: 1) Express loga X2 + 2 loga Square root of X as a single logarithm. Simplify if possible.
2) Given logb 2=0.387 and logb 5=0.898 find log square root of 20
3) solve 2x+3
5 =125
4)logx +log(x-2)=1

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1) Express loga X^2 + 2 loga Square root of X as a single logarithm. Simplify if possible.
=2logX + 2log a = 2log(Xa)
2) Given logb 2=0.387 and logb 5=0.898 find log square root of 20
log20 = log(4*5) = 2log2 + log5 = 2(0.387)+0.898=1.672


3) solve 2x+35 =125
2x=90
x=45
4)logx +log(x-2)=1
log[x(x-2)]=1
log[x^2-2x]=1
x^2-2x=10
x^2-2x-10=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-2x%2B-10+=+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=%28-2%29%5E2-4%2A1%2A-10=44.

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

x%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+44+%29%29%2F2%5C1+=+4.3166247903554
x%5B2%5D+=+%28-%28-2%29-sqrt%28+44+%29%29%2F2%5C1+=+-2.3166247903554

Quadratic expression 1x%5E2%2B-2x%2B-10 can be factored:
1x%5E2%2B-2x%2B-10+=+1%28x-4.3166247903554%29%2A%28x--2.3166247903554%29
Again, the answer is: 4.3166247903554, -2.3166247903554. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-10+%29

Cheers,
Stan H.