Tutors Answer Your Questions about Exponential-and-logarithmic-functions (FREE)
Question 80500: I can't understand how exactly to solve this.
the problem is:(2xy)^2(4x^2y^3)^2.
I don't know if I was on the right track or not but I started off by squaring everything and then didn't know what to do after that.
Click here to see answer by checkley75(3666)  |
Question 80522: Solve for x: log(base 5)3x + log(base 5)(x-3) = 1
I see they both have a common base of 5. I'm not sure if I'm on the right track, but this is what I got so far:
log(base 5)[3x(x-3)] = 1
3x(x-3) = 5^1
3x^2 - 9x =5
3x^2 - 9x - 5 = 0
I then tried factoring it out, but I don't think it works.
I think that first 3 in the problem might be throwing me off for some reason. Any help with this is greatly appreciated. Thanks.
Click here to see answer by Earlsdon(6294) |
Question 77618: i need some help can some please help me .And thanks in advance .
How can translate to algebra statement ;do not solve :
Three times the difference of five and a twice a number yields the same result as the same number increased by four .
Thanks for helpimg me .
Click here to see answer by rajagopalan(158)  |
Question 80626: A colony of bacteria has 6.5 X 10^6 members at 8 A.M. and 9.75 X 10^6 members at 10:30 A.M. Find its population at noon.
I used the formula N = No X 2 ^ t/d
9.75 X 10^6 = 6.5 X 10^6 X (2)^ 2.5/d
how do i work from here?
Click here to see answer by stanbon(57966) |
Question 80635: If 40mg of a radioactive substance dcays to 5mg in 12min, find the half-life, in minutes of the substance.
The halflife formula is
N = No (1/2) ^t/h
so i think that the formula goes to
5 = 40 (1/2) ^ 5/h but im not sure if thats right and i can't figure out how to solve
Click here to see answer by stanbon(57966) |
Question 80870: Find an antiderivative of the function
(7 *e^(2 *x ))/(5+e^(4*x))
involving the arctan function.
Note. Powers of e may be input either directly or using the function exp, e.g. e2 may be input as e^2 or exp(2). There is an equivalent solution to the problem in terms of ln (or log). However, the solution is required to be in terms of the arctan function (i.e. the inverse of the tan which is sometimes written as tan-1). Please enter arctan and not tan-1.
Click here to see answer by kev82(148) |
Question 81149: The air pressure P at sea level is about 14.7 pounds per square inch. As the altitude h(in feet above sea level) increases, the air pressure decreases. The relationship between air pressure and altitude can be modeled by
P=14.7e^-0.00004h
Mount everest in Tobet and Nepal rises to a height of 29,028 feet above sea level. What is the air pressure at the peak of Mount Everest??
Click here to see answer by stanbon(57966) |
Question 81530: Here is the problem I am having trouble with:
-Find the solution of the exponential equation, correct to four decimal places.
5^x = 4^x+1
so far I got:
ln5^x = ln4^x+1
(x)ln5 = (x+1)ln4
(x)ln5 = xln4 + ln4
If this is right, I'm not sure how to isolate x at this point.
Click here to see answer by rapaljer(4667)  |
Question 81536: Hello:
Here is my problem that I need assistance with. Thank you.
The period T (in seconds) of a simple pendulum is a function of its length L (in feet), given by T(L) = 2 pi square root of L/G where G = 32.2 feet per second. Per second is the acceleration of gravity. Express length L as a function of the period T.
Click here to see answer by tutor_paul(490)  |
Question 81739: The problem: Solve the equation.
x^2(2x)-2^x=0
This is how I'm thinking it starts:
x^2(2x)-2^x=0
2x^3-2^x=0
2(x^3-1^x)=0
x^3-1^x=0
3lnx=xln
If this is right, I'm not sure where to take it at this point.
Click here to see answer by rapaljer(4667) |
Question 81826: If I had the equation, 60.6=58e^-k(1), how would I solve it using Newton's Law of Cooling? I am clueless when it comes to Newton's Law of Cooling, so could you please explain that to me as well. Thank you in advance.
Click here to see answer by stanbon(57966) |
Question 81809: A Story Problem:
My country is interested in developing its natural resources to better the lives of its people. Unlike many developing countries, we wish to develop our resources in a sustainable way. Our glorious president the Honorable Cesa Mucumboo commissioned a nationwide survey of the country in 2006 to ascertain what minerals, forests and other resources were available. Unfortunately, none of the resources discovered were sustainable or feasible for the long term prosperity of Filam.
In reviewing the data from this survey, I noted that several geologist collected rocks which they described as “very old”. I feel that the location at which these rocks were collected could become a tourist destination if they prove to be some of the oldest rocks on the planet.
To help us determine the age of these rocks, I sent a sample of overlying rock to the University of Tunzaboo for testing. The geoscientists dated the rocks as only being 2.1 billion years old (compared to the oldest rocks at 4.25 billion years old.). Since the prosperity of our nation depends on the accuracy of these results, I need a second opinion on dating these rocks.
Potassium-argon dating was used to obtain the estimate of these rocks. Potassium-argon dating is based on the idea the potassium 40 decays into Argon 40 at a known rate. This rate is based on the half life of potassium 40 which is 1.25 billion years. This means that an original amount of potassium 40 decays by half into argon 40 every 1.25 billion years according to
y=y0e^-kt
In this equation, y0 is the original amount of potassium 40 and y is the amount of potassium 40 at some later time t in years. The constant k is found using the half life of potassium 40.
Using an independent lab, a sample of overlying volcanic rock was found to contain 52.4 micrograms of Potassium 40 and 153.6 micrograms of Argon 40. Assume that all of the argon 40 was originally potassium 40. Based on this data and the equation above, I would like you to determine the age of the overlying rock. The underlying rock must certainly be older.
Click here to see answer by scott8148(6628)  |
Question 81933: Here is the offending problem:
6e^5t = 18e^3t+2
I'm not sure how to start this one. My professor didn't give us an example like this that I can reference. Here is my attempt:
6e^5t = 18e^3t+2
6e^5t - 18e^3t+2 =0
6(e^5t - 2e^3t+2) = 0
At this point I am lost.
Click here to see answer by scott8148(6628) |
Question 82632: I am having a real difficulty in figuring out how to solve these exponential equations. Your help is greatly appreciated.
Solve each exponential equation. Give answers to four decimal places when necessary:
2^x^2-2x =8 and 3^x^2+4x = 1/81
Click here to see answer by josmiceli(9817)  |
Question 83066: Hi, help! I clicked on the link for the formula plotting system but it takes me to a wiki page with a huge list of pharmaceuticals...! Here's the problem, please note that the exponent says .75 but it should read 3/4, I just couldn't figure out how to do it so it would come out right.
For the function f, defined above, what are all the values of x for which f(x) is a real number?
Click here to see answer by stanbon(57966) |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580
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