Please help me with the following:?

Question

Solve for x: log5 x + log5 (x-4) = log5 12

is the answer: a) 4 b) 5 c) 6 d) 7

please list the steps for solving..
thank you

Answer 1

Since the log of products is the same as the sum of logs, you can also turn the sum of logs into the log of a product: log5x+log5(x-4)=log5[x(x-4)]=log512
Now this problem is reduced to x(x-4)=12
or x^2-4x=12
x^2-4x-12=0
(x-6)*(x+2)=0
x=6,-2
However, x=-2 does not qualify since it is not in the domain of log5x. Consequently, x=6 is the only solution.

Answer 2

log(x) + log(y) = log(xy), so this is

log5 (x(x - 4)/12) = 0, raise 5 to both sides:

x(x-4)/12 = 1, x^2 - 4x - 12 = 0, (x - 6)(x + 2) = 0
x = 6 or x = -2

BUT... you must check your answer for extraneous results. For x = 6

log5(6) + log5(2) = log5(12) : check

for x = 2

log5(2) + log5(-2) = log5(12) : log5(-2) is undefined, so the only answer is 6

Answer 3

I'll omit the base for brevity. (i.e. log A means logarithm of A to the base 5)

so you're asking for the solution of the equation:

log x + log ( x - 4 ) = log 12

recall that log A + log B = log ( AB ), hence you may rewrite the left-hand side of your original equation as:

log x + log ( x - 4 ) = log [ x(x - 4) ] = log 12

since the logarithmic function is a one-to-one function, log C = log D implies C = D; hence,

x(x - 4) = 12
x^2 - 4x = 12
x^2 - 4x - 12 = 0
(x - 6)(x + 2) = 0
x - 6 = 0 OR x + 2 = 0
x = 6 OR x = -2

but x cannot be equal to -2 since x appears as an argument for the logarithmic function in log x (i.e. the first term in the left-hand side of the original equation).

let us check whether x = 6 is indeed a solution:

log [x(x - 4)] = log [6(6 - 4)] = log [6(2)] = log 12, which checks!

Answer 4

log5 x+log5 (x-4) =log5 12
log5 [x(x-4)]=log5 12
x(x-4)=12
x^2-4x-12=0
(x-6)(x+2)=0
x=6 or x=-2(NA)