#1
can anybody help..
Solve for x:

(5 exponent x+3) - 3(5 exponent x+2) + (5 exponent x+1)= 6875

and find the zeroes of


f(x) = [5 exponent x(x-6)] - 1/625

i try consulting the math thread but after an hour i got no replies so sorry for breaking the rules...

and thanks in advance
#2
If no one in the math thread can help you, I doubt the Pit can. Regular posters check the 'ONLY' threads regularly. Just be patient. It's still on the front page!
#3
changes 6875 to 11 (5^4) then you can ingnore all the fives and solve the equation. x+3 -3(x+2)+ x +1= 11(4). Im not 100% sure you can do that but it makes sense to me.
#4
in future... exponent is written like 5^(x+3)

problem 1.

5^(x+3) = 5^(x) . 5^(3) = 125. 5^x

do it for all of them...

125.5^x - 3(25.5^x) + 5.5^x= 6875
expand...

125.5^x - 75.5^x + 5.5^x= 6875

solve... 55.5^x = 6875

5^x = 125

Log of base 5 each side and u get x=3


The 2nd problem take a log of base 5 of both sides...

x(x-6) = -4 (since 5^-4 = 1/625)

x^2 -6x + 4 = 0

kgo solve.
If looks coud really kill,

then my profession would be staring.

#7
Quote by MasterStafy
....geez


yeah i know what u mean
i solve his problem and not even get acknowledged?
pffft
If looks coud really kill,

then my profession would be staring.