Find x for lg2 + lg[ 4^(x-2) + 9 ]= 1 + lg[ 2^(x-2)+ 1 ]

We'll determine the constraints of existance of logarithm
function.

4^(x-2) + 9 >
0

We'll subtract 9 both
sides:

4^(x-2) >  9

For
any value of x, 4^(x-2) > 0.

2^(x-2)+ 1 >
0

We'll subtract 1 both
sides:

2^(x-2)>-1

For
any value of x, 2^(x-2) > 0.

Now, we'll solve the
equation.

We notice that, to the left side, we have the sum
of 2 logarithms that have the same base. We'll use the product property of logarithms
(the sum of logarithms is the logarithm of the
product).

lg2 + lg[ 4^(x-2) + 9 ]=lg {2*[ 4^(x-2) + 9
]}

We'll re-write the value 1, from the right side, using
the decimal logarithm.

1 = lg
10

We'll re-write the right
side:

1 + lg[ 2^(x-2)+ 1] = lg 10 + lg[ 2^(x-2)+
1]

We'll also transform the sum into a
product:

lg 10 + lg[ 2^(x-2)+ 1] = lg 10*[ 2^(x-2)+
1]

We'll re-write the entire
equation:

lg {2*[ 4^(x-2) + 9 ]} = lg {10*[ 2^(x-2)+
1]}

Since the bases are matching, we'll apply the one to
one property:

2*[ 4^(x-2) + 9] = 10*[ 2^(x-2)+
1]

We'll remove the
brackets:

2*4^(x-2) + 18 = 10*2^(x-2) +
10

We'll write 4^(x-2) as 4^x / 4^2, 2^(x-2) as 2^x / 2^2
and we'll subtract 18 both sides:

2 * 4^x / 4^2 = 10*2^x /
2^2 + 10 - 18

4^x / 2^3 = 5*2^x / 2 -
8

We'll move all terms to one
side:

4^x / 8 - 5*2^x / 2 + 8 =
0

We'll bring all ratios to the common denominator,
8.

4^x - 5*4*2^x + 64 =
0

We'll note 2^x = t.

t^2 -
20t + 64 = 0

We'll apply the quadratic
formula:

t1 = [20 + sqrt(400 - 256)] /
2

t1 = (20+12)/2

t1 =
32/2

t1 = 16

t2 =
(20-12)/2

t2 = 8/2

t2 =
4

2^x = t1

2^x =
16

2^x = 2^4

x =
4

2^x = t2

2^x
= 4

2^x = 2^2

x
= 2

Since both solutions are
positive, they are both valid.