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

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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**

For any positive value of k, log (a) b = log (k) b / log (k) a

log (16) (x^2 + 14x – 2) = log (4) (x – 1)

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

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

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

=> log (4) (x^2 + 14x – 2) = 2 * log (4) (x – 1)

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

take the antilog of both the sides

=> (x^2 + 14x – 2) = (x – 1) ^2

=> (x^2 + 14x – 2) = (x^2 + 1 – 2x)

=> x^2 + 14x – 2 - x^2 - 1 + 2x =0

=> 16x – 3 =0

=> x = 3/16

**Therefore x is equal to 3/16.**

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

The left side is the logarithm of base 16 and the right side is the logarithm of base 4. So we change the the base of left side also to 4 as below:

log (16) y = log (4) y/log(4)16 = log4(y)}/log(4) 4^2 ={log(4)y}./2 = (1/2)log(4) y = log(4) y^(1/2)

So we rewrite the equation as below:

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

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

We take antilog with respect to base 4 on both side:

{x^2+14x-2)^(1/2) = (x-1).

We square both sides:

{x^2+14x-2) = (x-1)^2 = x^2-2x+1.

{x^2+14x-2) - (x^2-2x+1) = 0

16x-3 = 0.

16x = 3.

x= 3/16.