How to solve these log functions

7. log (x) + log (x-15) = 2

8. log5 m = 1 + log5 (m+ 2)

9. log(x+9) = log4x

10. -3log3(-9x) = -3

log(x) + log(x - 15) = 2
log(x * (x - 15)) = 2
x * (x - 15) = 10^2
x^2 - 15x = 100
x^2 - 15x - 100 = 0
x = (15 +/- sqrt(225 + 400)) / 2
x = (15 +/- sqrt(625)) / 2
x = (15 +/- 25) / 2
x = 40/2 , -10/2
x = 20 , -5

x = 20 is the only solution that works in the original equation


log[5](m) = 1 + log[5](m + 2)
log[5](m) - log[5](m + 2) = 1
log[5](m / (m + 2)) = 1
m / (m + 2) = 5^1
m / (m + 2) = 5
m = 5m + 10
-10 = 5m - m
-10 = 4m
m = -5/2


log(x + 9) = log(4x)
x + 9 = 4x
9 = 3x
3 = x


-3 * log[3](-9x) = -3
log[3](-9x) = 1
-9x = 3^1
-9x = 3
x = -3/9
x = -1/3

for the first one, because of the properties of logs, we can multiply what is inside of the parentheses and make it one big log.

log x^2 - 15x = 2

Logs without a specified base are base ten, which means that the equation can be rewritten as

10^2 = x^2 - 15x

and factored gives us

(x-20)(x+5)=0

Because we cant have a negative number logged, the answer is x = 20

for the second one

rewrite the equation as:

log5m - log5(m+2) = 1

due to our properties of logs, we can rewrite the equation with one big log as:

log5(m/(m+2)) = 1

which can be rewritten as

5 = m/(m+2)

m = -5/2

for the third one

Set x + 9 = 4x

x = 3

for the fourth one

divide both sides by -3 to get

log3(-9x) = 1

Can be rewritten as

3 = -9x

x = -1/3

7- log(x)*(x-15)=2 log (x^2-15x)=2 x^2-15x =10^2 x^2-15x-100=0 (x+5)(x-20)=0 x= -5 or x=20
x=-5 will be rejected

8- log5 m - log5 (m+2)=1 log5 m/m+2 =1 m/m+2=5^1 m=5(m+2) m=5m+10 m=-10/4

9- x+9=4x 9=3x x=3