Question

Solve the logarithmic equation. Be sure to reject any value that is not in the domain of the original logarithmic expressions. Give the exact answer.

log6x2 = log6(5x + 36)

A. (3/2)

B. {9}

C. Ø

D. {9, -4}

Answer 1

log 6 + logx^2 = log6 + log(5x+36)

logx^2 = log(5x+36)

x^2 = 5x + 36
lol ....I am stuck after that

Answer 2

Well I'll continue where you left off:
x^2 = 5x + 36
x = 5x(1/2) + 36(1/2)
(You have to multiply 1/2 to get rid of the exponent)
x = 2.5x + 18
1.5x = 18
x = 12

I think that's right . . .

Answer 3

\[x ^{2}-5x-36=0\]
1*-36=-36
9-4=5
9*-4=-36
\[x ^{2}-\left( 9-4 \right)x-36=0\]
factor and find the values of x,regect any negative value.

Answer 4

@surjithayer the equation u started off with is wrong
This is the correct equaton\[x ^{2}-5x+36=0\]

Answer 5

i am not sure though

Answer 6

\[x ^{2}-9x+4x-36=0\]
x(x-9)+4(x-9)=0
(x-9)(x+4)=0
It gives x=9,-4
solution is {-4,9}
No need to reject Because(-4)^2=16,which is positive
Also 5*-4+36=16 which is positive

Answer 7

The equation you solved is
\[x ^{2}=5x+36\]

Answer 8

ohhh