# Math, Precalc

posted by Susan

(3^x)+(5^{x+3})=(3^{x+4})-(1/3)*( 5^{x+2}).

Solve

1. Anonymous

3^x + 5^(x+3) - 3^(x+4) + (1/3)5^(x+2)= 0
3^x(1 - 3^4) + 5^(x+2)(5 + (1/3)(1) ) = 0
-80(3^x) + (16/3)(5^(x+2) = 0
times -3/16
15(3^x) - 5^(x+2) = 0
15(3^x) = 5^(x+2)
divide by 5
3(3^x) = 5^(x+1)
3^(x+1) = 5^(x+1)
this can only be true if x = -1
that is, if 3^0 = 5^0 , which is true

2. Reiny

3^x + 5^(x+3) - 3^(x+4) + (1/3)5^(x+2)= 0
3^x(1 - 3^4) + 5^(x+2)(5 + (1/3)(1) ) = 0
-80(3^x) + (16/3)(5^(x+2) = 0
times -3/16
15(3^x) - 5^(x+2) = 0
15(3^x) = 5^(x+2)
divide by 5
3(3^x) = 5^(x+1)
3^(x+1) = 5^(x+1)
this can only be true if x = -1
that is, if 3^0 = 5^0 , which is true

3. Steve

just start by combining like powers:

(3^x)+(5^{x+3})=(3^{x+4})-(1/3)*( 5^{x+2})
3^x + 125*5^x = 81*3^x - 3^-1 * 25*5^x

3*3^x + 375*5^x = 243*3^x - 25*5^x
(375+25)*5^x = (243-3)*3^x
400*5^x = 240*3^x
5*5^x = 3*3^x
5^(x+1) = 3^(x+1)
(x+1)log5 = (x+1)log3
(x+1)(log5-log3) = 0
x = -1

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