Math, Precalc
posted by Susan
(3^x)+(5^{x+3})=(3^{x+4})(1/3)*( 5^{x+2}).
Solve

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 
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 
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 = (2433)*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)(log5log3) = 0
x = 1
Respond to this Question
Similar Questions

Math
Precalc stuff.. square root of z1= square root of z +6 are you solving for z? 
PreCalc
Solve for the variable m^2 + 2m +3 > = 0 
precalc
1. Solve 2x + 5< 9 
Math Precalc
could someone please help me solve this problem... e^x  12e^(x)  1 
precalc
solve x/x^24  1/x+2=2 
Math  PreCalc (12th Grade)
Use the inverse matrix to solve this system of equations: 4x+3y=7.5 7x+9z=14 4yz=8.3 4,3,0 7,0,9 0,4,1 
PreCalc
Solve each inequality: (1/(2b+1)) + (1/(b+1)) > (8/15) (2/(x+1)) < 1  (1/(x1)) My teacher told me the answer is in interval notation, and that a number line is used to solve. 
precalc
i also need help with these two questions: 1/r=1/r1 +1/r2 solve for r1 and: f= G(mM/r^2); solve for m 
precalc
(2)^x =  1/128 solve for x 
PreCalc
Solve: (x1)/(x)+(x2)/(x)+(x3)/(x)+...+(1)/(x)=3; X belongs to N