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

we know that log 10 = 1

We know that log a + log b = log a*b

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

We know that if log a = log b, ==> a = b

==>...

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

we know that log 10 = 1

We know that log a + log b = log a*b

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

We know that if log a = log b, ==> a = b

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

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

==> 2*4^(x-2) + 8 = 10*2^(x-2)

==> 2*[2^(x-2)]^2 = 10*2^(x-2) + 8 = 0

Assume that 2^(x-2) = y

==> 2y^2 - 10y + 8 = 0

==> y^2 - 5y + 4 = 0

==> (y-4)(y-1) = 0

==> y1= 4 ==> 2^(x-2) = 4 ==> x-2 = 2 ==> **x1 = 4**

==> y2= 1 ==> 2^(x-2) = 1 ==> x-2 = 0 ==> **x2= 2**