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3. Solve this exponential equation for x: e^5x=1/e^2x+7

4. Solve this exponential equation for x: 2^4x-7=32

I really need the value for x on each. Looking at my book didn't help, just left me cock-eyed.

2006-07-13 15:48:00 · 5 answers · asked by samadhi97 1 in Science & Mathematics Mathematics

5 answers

take the log of each side, then pull the exponent out front of the equation and solve algebraically.

2006-07-13 15:54:59 · answer #1 · answered by SnowXNinja 3 · 0 1

For the first one:

e^5x = e^-(2x+7)
ln (e^5x) = ln (e^-(2x+7))
5x = -(2x+7)
5x + 2x = -7
7x = -7
x = -1

For the second one:

2^(4x-7) = 32
2^(4x-7) = 2^5
log2(2^(4x-7)) = log2(2^5) note: log2 is log base 2
4x-7 = 5
4x = 12
x = 3

2006-07-13 23:01:53 · answer #2 · answered by SkyWayGuy 3 · 0 0

3)
e^5x = 1 / e^2x+7
First realize that 1 / e^2x+7 = e^-(2x+7)


When you have the same bases, you can equate the exponents

e^5x = e^-(2x+7)

So,
5x = -2x - 7
7x = 7
x = 1


4)
2^(4x-7) = 32

32 = 2^5

So, 2^(4x-7) = 2^(5)

4x - 7 = 5
4x = 12
x = 3

2006-07-13 22:57:33 · answer #3 · answered by Anonymous · 0 0

solution 1:
e^5x = 1 /(e^2x+7)
e^5x . e^(2x+7) = 1
e^(5x+2x+7) = 1
e^(7x + 7) = 1
len e^(7x+7) = ln 1
7x + 7 = 0
x = -1

solution 2:
2^(4x-7) = 32
2^(4x - 7) = 2^5
4x - 7 = 5
4x = 12
x = 3

i hope this can help

2006-07-13 23:25:24 · answer #4 · answered by Anonymous · 0 0

take the natural log of both sides. It will bring the x's down, then you'll have easy equations to deal with

2006-07-13 22:53:54 · answer #5 · answered by darcy_t2e 3 · 0 0

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