Please help me solve for x . the x + 2 for the 32 is an exponent and there weren't originally parentheses around it, i just did it to make it more clear...
Please show steps and explain, i appreciate the effort !
32^(x+2) = 4^x Solve for X?
2008-12-04 6:48 am
回答 (6)
2008-12-04 6:57 am
✔ 最佳答案
The trick is to find a common base. In general, if we have x^a = y^b; then if x = y, then a = b. i.e. if bases are equal then the exponents are equal.Note: 2^5 = 32 and 2^2 = 4:
2^(5 * (x + 2) ) = 2^(2*x)
We have equal bases so we can equate the exponents:
5 * (x + 2) = 2x
5x + 10 = 2x
10 = -3x
x = 10/-3 = -(10/3)
And that's our final answer. Hope this helps.
2008-12-04 6:19 pm
32^(x + 2) = 4^x
(2^5)^(x + 2) = (2^2)^x
2^[5(x + 2)] = 2^(2x)
5(x + 2) = 2x
5*x + 5*2 = 2x
5x - 2x = -10
3x = -10
x = -10/3
(2^5)^(x + 2) = (2^2)^x
2^[5(x + 2)] = 2^(2x)
5(x + 2) = 2x
5*x + 5*2 = 2x
5x - 2x = -10
3x = -10
x = -10/3
2008-12-04 3:11 pm
32=2^5,4=2^2
2^(5(x+2)=2^2x
5x+10=2x
3x=-10
x=-10/3
2^(5(x+2)=2^2x
5x+10=2x
3x=-10
x=-10/3
2008-12-04 3:07 pm
Use logarithms.
Since log a^b = b log a
we have (x+2) log 32 = x log 4
32 and 4 are both powers of 2 32 = 2^5 and 4=2^2
So using the above rule, we have (x+2)*5*log2 = x*2*log 2
Divide both sides by log 2
(x+2)*5=2x
This is 3x+10=0
solving, x=-10/3
Since log a^b = b log a
we have (x+2) log 32 = x log 4
32 and 4 are both powers of 2 32 = 2^5 and 4=2^2
So using the above rule, we have (x+2)*5*log2 = x*2*log 2
Divide both sides by log 2
(x+2)*5=2x
This is 3x+10=0
solving, x=-10/3
參考: confirmed with HP50g calculator
and graphically with WZGrapher
2008-12-04 2:58 pm
My brain hurts.
2008-12-04 2:58 pm
(2^5)^(x+2)=(2^2)^x , 2^(5x+10)=2^2x,
5x+10=2x , 3x=-10,
x=-10/3
5x+10=2x , 3x=-10,
x=-10/3
收錄日期: 2021-05-01 11:33:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081203224830AAtsDi7