Can´t solve logs?

Question

Hi.
Can someone please help me with the following problems?
1) log (2x^2 + 11x +5) - log (2x+1) = 2
2) log sqrt x^2 +1999 = 3
Can someone please walk me through this step by step?
Thank you!

Answer 1

You know the basics of Laws of Logarithms, right?
1) log (2x^2 + 11x +5) - log (2x+1) = 2
log [(2x^2 + 11x +5)/(2x+1)] = log 10^2
2x^2 + 11x +5 = 100(2x +1)
2x^2 + 11x +5 - 200x -100 = 0
2x^2 -189x -95 = 0
(2x+1)(x-95) = 0
x= -0.5 or x = 95

2) log [sqrt (x^2+1,999)] = 3 = log 10^3
sqrt (x^2 +1,999) = 1,000
x^2 + 1999 = 1,000,000
x^2 = 998,001
x = +/- 999

Answer 2

1) combine the two logs: log[ (2x^2+11x+5)/(2x+1)]=2
Un-log both sides: (2x^2+11x+5)/(2x+1)=10^2
Factor the top of left side: 2x^2+11x+5 becomes (2x+1)(x+5)
Simplify the left side by cancelling the 2x+1. What's left is
x+5 = 100, which gives x=95

2) Where are the parenthesis in this one?
for log(sqrt(x^2+1999))=3, unlog both sides:
sqrt(x^2+1999)=1000, square both sides
x^2+1999 = 1,000,000 solve
x=±999

Answer 3

(a) log(m*n) = log(m) + log(n)
(b) log(m^n) = n * log(m)
(c) If [log(m) = n] m = b^n [where b is the base, say 10]

1) Simplify (2x^2 + 11x +5) == (2x+1)(x+5)
apply (a) to get log (2x^2 + 11x +5) = log(2x+1) + log(x+5)
Simplifying further you get
log(x+5) = 2 ... apply (c) and find x

2) sqrt(x^2 + 1999) == (x^2 + 1999)^0.5
Apply (b) then (c) to solve for x

Answer 4

1.
log (2x^2 + 11x + 5) - log (2x+1) = 2
Factorise quadratic:
log[(2x + 1)(x + 5)] - log(2x + 1) = 2
Use log(a) - log(b) = log(a/b):
log[ (2x + 1)(x + 5) / (2x + 1)] = 2
Simplify fraction by cancelling 2x + 1:
log(x + 5) = 2
Use number = base^log, taking k as base of the logs:
x + 5 = k^2
Subtract 5:
x = k^2 - 5.

2.
log (sqrt (x^2 +1999)) = 3
Use number = base^log, taking k as base of the logs:
sqrt(x^2 + 1999) = k^3
Square:
x^2 + 1999 = k^6
Subtract 1999:
x^2 = k^6 - 1999
Take square root:
x = +/-sqrt(k^6 - 1999).

If you know the base of the logs, (say e or 10) substitute that for k.

Answer 5

Assume all logs are log base 10:-

Question 1
log [ (2x + 1) (x + 5) / (2x + 1) ] = 2
log (2x + 5) = 2
x + 5 = 10²
x = 95

Question 2
log [ (x² + 1999)^(1/2) ] = 3
(1/2) log (x² + 1999) = 3
log (x² + 1999) = 6
x² + 1999 = 10^6
x² = 10^6 - 1999
x = ± 998,001