does anyone know how to solve this log; log x + log(x+4)=1?

Answer 1

log x + log(x+4)=1

make left side into single expression
log x(x+4)=1

Apply the inverse log base 10 to both sides.
10^(log x(x+4)) = 10^1

simplify
x(x+4) = 10^1 = 10
x^2 + 4x = 10
x^2 +4x -10 =0

solve for x
x = -(√14 + 2) or x=(√14 - 2)

but since x = -(√14 + 2) is undefined for the original problem,
only x=(√14 - 2) is a valid answer.

Checking
log x + log(x+4)=1
log (√14 - 2) + log(√14 - 2 +4) = 1
log (1.7416574) + log(5.7416574) = 1
(0.2409627) + (0.7590373) = 1
1 = 1 check


see http://www.math.ucalgary.ca/~lpbos/math253/f06/examples/week1.pdf
for more examples with logs

Answer 2

By the Law of Logarithms:

log a + log b = log ab

In this case a = x and b = x+4

log x + log (x+4) = log (x^2 + 4x)

log (x^2 + 4x) = 1

x^2 + 4x = 10^1 = 10

x^2^ + 4x -10 = 0

Now we have a quadratic equation that can be solved to find the values of x.

Answer 3

Log (x(x+4)) = 1, or

x(x+4) = b, where b is the base of the logarithm. Solve for x.

Answer 4

since log a + log b = log ab
then
log x + log (x+4) =1
=> log [x/(x+4)] = 1
=> x = e(x+4)
=> x(1-e) = 4e
=> x = -4e / (e-1)