2007-03-28 15:23:39 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Treating this as solving for x where 1 + 5^(2x) = 7^(3x) -2, this is akin to 5^2x = 7^(3x) - 3.. It's easy to show that this equation can't be true for any x => 1, because the right side would be much larger than the left side. If x = 0 then the left side is 1 and the right side is -2, making the left side larger than the right side. The answer is therefore between 0 and 1. (It's actually about 0.2945.)
2007-03-28 15:44:51 · answer #1 · answered by Isaac Laquedem 4 · 0⤊ 0⤋
just use parentheses
5^(2x+1) = 7^(3x-2)
take log for both sides
log5^(2x+1) = log7^(3x-2)
power property, which you move the exponent infront of log
(2x+1)log5 = (3x-2)log7
distribute
2xlog5 + log5 = 3xlog7 - log49
sutract 2xlog5 for both sides
log5 = 3xlog7 - 2xlog5 - log49
add log49 for both sides
log5 + log49 = 3xlog7 - 2xlog5
product property (left), and take out x (right)
log(5*49) = x (3log7 - 2log5)
power property
log245 = x (log343 - log25)
quotient property
log245 = x(log343/25)
x = log245 / log(343/25)
x = 2.100635203
2007-03-28 22:35:15 · answer #2 · answered by 7 · 0⤊ 0⤋
Take natural log of both sides:
(2x+1)ln(5) = (3x-2)ln(7)
Then solve for x:
2xln(5) + ln(5) = 3xln(7) - 2ln(7)
ln(5) + 2ln(7) = 3xln(7) - 2xln(5)
ln(5) + 2ln(7) = x(3ln(7) - 2ln(5))
x = (ln(5) + 2ln(7)) / (3ln(7) - 2ln(5))
x = 2.1006
2007-03-28 22:39:02 · answer #3 · answered by CheeseHead 2 · 0⤊ 0⤋
You need to tell us with parentheses where the exponents end.
Take logs of both sides, and use the fact that LOG(a^b)=bLOG(a).
2007-03-28 22:36:39 · answer #4 · answered by fcas80 7 · 0⤊ 0⤋
zero
2007-03-28 22:31:32 · answer #5 · answered by Anonymous · 0⤊ 0⤋