SOLUTION: how can i solve logX + logX + 1 = log12 ?

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: how can i solve logX + logX + 1 = log12 ?      Log On


   



Question 132012: how can i solve
logX + logX + 1 = log12 ?

Found 2 solutions by josmiceli, rapaljer:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
log%28X%29+%2B+log%28X%29+%2B+1+=+log%2812%29
First get everything in terms of logs
1 = log(x)
If the base is 10, then I can write
10%5E1+=+x
x+=+10, so
1+=+log%2810%29
Now rewrite it
log%28X%29+%2B+log%28X%29+%2B+log%2810%29+=+log%2812%29
The general rule for logs is:
log%28a%2Ab%2Ac%29+=+log%28a%29+%2B+log%28b%29+%2B+log%28c%29 so,
log%28X%29+%2B+log%28X%29+%2B+log%2810%29+=+log%28X%2AX%2A10%29
log%2810x%5E2%29+=+log%2812%29
If the logs are equal, the numbers are equal
10X%5E2+=+12
X%5E2+=+6%2F5
X+=+sqrt%286%2F5%29 answer
There is a negative square root also, but it makes
no sense here, since there is no log to the base 10
that will give me a negative number
check:
sqrt%286%2F5%29+=+1.0954
log%28X%29+%2B+log%28X%29+%2B+1+=+log%2812%29
log%281.0954%29+%2B+log%281.0954%29+%2B+1+=+log%2812%29
.03957+%2B+.03957+%2B+1+=+1.0792
.07914+%2B+1+=+1.0792
1.07914+=+1.0792
close enough

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
logX + logX + 1 = log12
2logX+1=log12
1=log12-2logx
1=log12-logx^2
1=log%2812%2Fx%5E2%29+
1=log%2810%2C12%2Fx%5E2%29

By basic definition of logarithms:
10%5E1=+12%2Fx%5E2
10x%5E2+=+12
x%5E2=12%2F10=6%2F5

Since x must be a positive number,
x=+sqrt%286%2F5%29
x=sqrt%2830%29%2F5

R^2