logarithms?

Question

log3 (5-x) + log3 x = log3 (8-x)

(that means base 3)
solve.. please show steps

Answer 1

log3 (5-x) + log3 x = log3 (8-x)
<=> log3 [(5-x).x] = log3 (8-x)
<=> (5-x).x = 8-x
<=> 5x - x^2 = 8 - x
<=> x^2 - 6x + 8 = 0
<=> (x-4)(x-2) = 0
<=> x = 2 or 4
We also need 5-x, x and 8-x to be strictly positive. In this case that doesn't rule out either solution, so the final answer is x = 2 or 4.

Answer 2

log[base 3](5 - x) + log[base 3](x) = log[base 3](8 - x)

Use the log identity to combine the logs on the left hand side.

log[base 3] ( (5 - x)x ) = log[base 3](8 - x)

Taking the antilog of both sides eliminates the logs, and we can equate the arguments.

(5 - x)(x) = 8 - x

Expand the left hand side,

5x - x^2 = 8 - x

Bring everything over to the right hand side,

0 = x^2 - 6x + 8

Factor,

0 = (x - 4)(x - 2)

Therefore, x = {4, 2}
However, because this is a logarithmic equation, we have to check for extraneous solutions.

Let x = 4:
LHS = log[base 3](5 - 4) + log[base 3](4)
= log[base 3](1) + log[base 3](4)
= 0 + log[base 3](4)
= log[base 3](4)
RHS = log[base 3](8 - 4)
= log[base 3](4)
Therefore, this solution works.

Let x = 2.
LHS = log[base 3](5 - 2) + log[base 3](2)
= log[base 3](3) + log[base 3](2)
= log[base 3](3*2)
= log[base 3](6)

RHS = log[base 3](8 - 2)
= log[base 3](6)
So x = 2 works.

Therefore, the solution is indeed x = {4, 2}

Answer 3

Assume "log" means log base 3:-
log ( (5 - x).(x) )= log (8 - x)
5x - x² = 8 - x
x² - 6x + 8 = 0
(x - 4).(x - 2) = 0
x = 2, x = 4

Answer 4

log(5-x) + log x = log (8-x)

log(x(5-x))=log(8-x)

x(5-x)=8-x (Since the bases are same)

5x-x*x=8-x

x^2-6x+8=0

(x-2)(x-4)=0

so x=2 or x=4