In the standard equation of an exponential function, why can't "b" be negative?
95797
Like fLoWeRs said....you didn't give the equation! So I'm going to ansswer you based on what I regard as a standard exponential.\n\nf(x) = ae^bx + c\n\nThis is a fairly standard exponential as far as I know, and there's no reason why "b" can't be negative. In fact, if it weren't in some cases, we'd be living in a pretty funky place.
648708
I know this! I did a problem like this a while ago! It's an exponential equation. It can be used to predict exponential increases or decreases, like bacteria or double/triple/quadrupled information.
1156569
The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.\n\nThe upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i,  pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.
653428
Because -2 squared = (-2)(-2)\n\nA negative times a negative equals a positive.
802801
i think -1; y=mx+b, "b" is the 3 and "m"(slope) is negative one.
814311
The issue lies in the fact that the original equation is 0=0.\n\nThen, if you put in any value for x, you are dividing by 0 which as we all know, is a "no no."
156464
u can't devide by (a-b) as (a-b)=0
1086154
The reason we call ln(x) "natural" is twofold: first, expressions in which the unknown variable appears as the exponent of e occur much more often than exponents of 10 (because of the "natural" properties of the exponential function which allow it to describe growth and decay behaviors), and second, because the natural logarithm can be defined quite easily using a simple integral or Taylor series--which is not true of other logarithms.
226422
This one is trickier - ln-2/x wasn't allowed on the source. I tried the following:\n\nln[e]-2/x ==> 1-2/x ==> 2/x^2\nln[2/x] ==> -1/x\n\nI hope this helps.
307451
Try\n\nhttp://www.purplemath.com/modules/negative.htm
520921
The (x-3)s cancel out because they will produce the same answer for all values of x.\n\n(x-5)/(5-x) produces an answer of -1 because they'll always be the opposite (negative) answer for all values of x.\n\nThe answer is -1
485166
The correct answer is d.\n\nf(1-x) will give a negative number for all positive 'x' values.\n\nbut squaring these negative number will always give a positive number i.e (-4)^2 = +16.\n\nTherefore in the equation f(x) = x^2(1-x)^2;\n\nYou can see that for all positive or negative numbers you will have a positive value. Again,\n\nLet x=5; Then 1-x = -4;\n\nFor the equation d.. 1-(-4) = 5. i.e the original number.\n\nHence, f(5) = 5^2 (-4)^2 = 25*16;\nAlso, f(-4) = (-4)^2 (5)^2 = 25*16;\n\nThus, for all values of x... Equation d is TRUE.
492387
This has no solution, 2^x would be equal to a negative number. That can't happen.
978128
log(base a) b = logb/loga.\n\nSo take the log of X and divide by the log of 1/3.  Then press the +/- button because you're being asked for the negative of that value.
909722
The only case that I could think of that A is not equal to B where AC=BC is when C=0.  Other than that A must equal B for the expression to be valid
985432
well...ln(x) does not cancel e^x....but...\n\ngiven e^x, if we take the natural log of e^x then we have ln(e^x), well this means that we have an exponent...we recall that we can pull the exponent down infront of a logrithm...so we now have xln(e)\n\nah! the ln(e) is 1, we recall that ln's are base e and if we take the logrithm of the base we get 1. so we have x*1=x\n\nso...likewise...lnx^2 is 2ln(x)...and thats all we can do\n\nhope this helps\n\nmatttlocke
622196
bc when u have 2 and you take 4 away from it, you are negative 2--- think of it like this: 2-4=-2