SOLUTION: Find a value for X in he equation 2log10X+3 = log10X+log10 500 i'm assured the second line is {{{ log10x^2 }}} +log10 1000 = log10X+log10 500 How do you get that from abo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Find a value for X in he equation 2log10X+3 = log10X+log10 500 i'm assured the second line is {{{ log10x^2 }}} +log10 1000 = log10X+log10 500 How do you get that from abo      Log On


   

Question 7157: Find a value for X in he equation
2log10X+3 = log10X+log10 500
i'm assured the second line is
+log10x%5E2+ +log10 1000 = log10X+log10 500
How do you get that from above, i tried using my calculator but it kept giving me errors.
Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
The second line is right. It's good to use the log property log a + log b = log (ab). So, the equation then would be:

+log%2810%2C1000x%5E2%29+=+log%2810%2C500x%29+ <---- Both logs are base 10, and so you can just set whatever you're finding the log of equal to each other:

+1000x%5E2+=+500x+

+1000x%5E2+-+500x+=+0+ <--- move 500x to the left

+500x%282x+-+1%29+=+0+ <---- Just by looking at this x=0 or x=1/2. We would throw out the x=0 because you can't find the log of 0. (If you have, for example +log%2810%2C0%29=x+, that would translate to the exponential form +10%5Ex+=+0+. And there is no such number for x that will make +10%5Ex+ zero.)