log問題（2條）（f.4程度）

1.
log2 (x + 5) - 3 = log2 (4 - x)
log2 (x + 5) = log2 (4 - x) + 3
log2 (x + 5) = log2 (4 - x) + 3 log22
log2 (x + 5) = log2 (4 - x) + log28
log2 (x + 5) = log2 8(4 - x)
x + 5 = 8(4 - x)
x + 5 = 32 - 8x
9x = 27
x = 3


2.
[3^(x+2)] - (3^x) - 72 = 0
(3^2)*(3^x) - (3^x) - 72 = 0
9(3^x) - (3^x) - 72 = 0
8(3^x) = 72
3^x = 9
3^x = 3^2
x = 2

3^(x+2)-3^x-72=0
3^x(3^2-1)-72=0
3^x(8)=72
3^x=9
3^x=3^2
x=2

lg2 ( X + 5 ) - 3 = lg2 ( 4 - X )

<=> lg2 ( X + 5 ) = 3 + lg2 ( 4 - X )

<=> lg2 ( X + 5 ) = lg2 ( 2^3 ) + lg2 ( 4 - X )

<=> lg2 ( X + 5 ) = lg2 [ ( 2^3 )( 4 - X ) ]

If { ( X + 5 ) / [ ( 2^3 )( 4 - X ) ] } and { [ ( 2^3 )( 4 - X ) ] / ( X + 5 ) } are defined,
then:

<=> ( X + 5 ) = [ ( 2^3 )( 4 - X ) ]

<=> ( X + 5 ) = ( 32 - 8X )

<=> 9X = 32 - 5

<=> X = 3 (ANSWER)
*********

3^(X+2) - 3^(X) - 72 = 0

<=> 9[3^(X)] - 3^(X) - 72 = 0

<=> 8[3^(X)] = 72

<=> 3^(X) = 9

Therefore X = 2 (ANSWER)