What are the values of x for log (16) (x^2 + 14x – 2) = log (4) (x – 1)?

log16 ( x^2 + 14x - 2) = log 4
(x-1)

First we will use the logarithm properties to
solve.

We will rewrite log
16

We know that log a b = log c b/ log c
a

==> log 16 ( x^2 +14x -2 ) = log 4 (x^2 +14x -2) /
log 4 16

                                           = log 4
(x^2 +14x -2) / log 4
4^2

                                           =  log 4
(x^2 +14x -2) / 2log 4 4

But log 4 (4) =
1

==> log 16 (x^2 +14x -2) = (1/2) log 4 (x^2 +14x
-2)

Now we will substitute into the
equation.

==> (1/2) log 4 (x^2 +14x -2) = log 4
(x-1)

==> log 4 (x^2 +14x -2)^1/2 = log 4
(x-1)

Now that the logs are equal, then the bases are
equal.

==> (x^2 +14x -2)^1/2 =
x-1

We will square both
sides.

==> x^2 +14x -2 = x^2 -2x +
1

We will reduce
similar.

==> 14x -2 = -2x
+1

We will combine like
terms.

==> 16x =
3

==> x =
3/16