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# Recent problems solved by 'jsmallt9'

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 logarithm/396144: log25-log125/3log51 solutions Answer 281288 by jsmallt9(3296)   on 2011-01-19 07:17:57 (Show Source): You can put this solution on YOUR website!Please repost your problem. Use parentheses around arguments to logarithms and use additional parentheses as needed to make your expressions/equations unambiguous. Here's some of the ways your expression could be interpreted: etc. Tutors are more likely to answer your questions if they are clearly stated.
 logarithm/396429: How do I simplify the following: 4 ^ log base 4 of 51 solutions Answer 281287 by jsmallt9(3296)   on 2011-01-19 07:05:09 (Show Source): You can put this solution on YOUR website!Once you understand what logarithms are this problem becomes very easy. The idea behind logarithms is that it is possible to take any positive number (other than 1) and, if you raise it to the correct power, get any other positive number. For example, it is possible to raise 12 to some power and get 1/2. Or raise 3 to some power and get 153. The tricky part is finding out wht these exponents are. What power of 12 results in 1/2? And what power of 3 results in 153? Logarithms are used to express these exponents. For example represents the exponent for 12 that results in 1/2. And represents the exponent for 3 that results in 153. Now let's look at Let's look at just the exponent: . This is "the exponent for 4 that results in 5" or "the power of 4 that results in 5". And where do we find ? Answer: As an exponent for 4! Since your expression raises 4 to the power of 4 that results in 5, your expression is simply a 5!
 Exponents-negative-and-fractional/396287: 3(x+1)^(4/3)=481 solutions Answer 281148 by jsmallt9(3296)   on 2011-01-18 16:41:27 (Show Source): You can put this solution on YOUR website! First let's isolate the base, (x+1), and its exponent, 4/3, on one side of the equation by dividing both sides by 3: Next we want to find a way to change the exponent from 4/3 to a 1. We can do so by combining all of the following ideas:One is allowed to raise both sides of an equation to any non-zero power.The rule for exponents when raising a power to a power is to multiply the exponents.Multiplying reciprocals always results in a 1. All of this tells us that if we raise each side of the equation to the reciprocal of 4/3 power, we will end up with an exponent of 1 on the left side. The reciprocal of 4/3 is 3/4: On the left side the exponent turns into a 1, as planned: Simplifying the left side is simple. The right side is not so simple. If you have trouble with negative or fractional exponents I find it can be helpful to factor the exponent in a certain way:If the exponent is negative, factor out a -1.If the exponent is fractional and if the numerator is not a 1, then factor out the numerator. (You'll see what this means in a moment.) The exponent on the right is not negative but it is a fraction whose numerator is not a 1. So we will factor out the 3: Now, looking at the exponent in factored form, the 3 tells us we will be cubing and the 1/4 tells us that we will be finding a 4th root. And we can do these two operations in any order! Normally 4th roots aren't simple. However, since , finding the 4th root of 16 is rather easy. So we will start with the 4th root and then cube it: x+1 = 8 Now we just subtract 1: x = 7
 Exponents-negative-and-fractional/395705: how do you even begin solving this problem (16 to the -5 power) to the 1/20 power1 solutions Answer 280839 by jsmallt9(3296)   on 2011-01-17 14:10:23 (Show Source): You can put this solution on YOUR website! As often happens there are several ways this problem can be done. The most obvious first step is to use the rule for exponents when raising a power to a power. The rule is to multiply the exponents: which simplifies as follows: Next we try to simplify this expression even more. If you have trouble with negative and/or fractional exponents, I find it helpful to factor the exponent is a certain way:If the exponent is negative, then factor out -1.If the exponent is fractional and the numerator is not a 1, then factor out the numerator. For example, if the exponent was 2/3, then factor out the 2 as follows: 2*1/3. After factoring the exponent in this way, look at the factors. Each one tells you one operation that will be done. And since multiplication is Commutative, these operations can be done in any order you choose! Let's see how this works on your expression. First we factor the exponent: Looking at the factors of the exponent, the -1 tells us that a reciprocal will be found. And the 1/4 tells us that a 4th root will be found. And the order we do them will not make any difference! So let's choose the order that seems the easiest. Normally a 4th root sounds difficult. But once you realize that 16 is then the 4th root is quite easy. So we will start with that. And then we will find a reciprocal. (Since 1/2 is the reciprocal of 2)
 Square-cubic-other-roots/395152: why is the answer to cube root of 9 times cubed root of 6 DIVIDED BY 6 root 2 times 6 root 2 = 3?1 solutions Answer 280542 by jsmallt9(3296)   on 2011-01-16 12:27:20 (Show Source): You can put this solution on YOUR website! First we can use a property of radicals, , to multiply in both the numerator and the denominator: or I left the radicand in the denominator in the form of for reasons that will become clear soon. Next we use fractional exponents in the denominator to help us simplify further. Using the pattern we can rewrite the denominator as: We can see that this fraction will reduce to: Writing this back in radical form we get: or Now our fraction looks like: Now we can use another property of radicals, to combine the two radicals into one: which simplifies to: And since , this cube root becomes a 3!
 Quadratic_Equations/395154: use the quadratic formula to solve the equation. If necessary, round answers to the nearest hundredth. question 1 : 2a^2-46a+252=0 question 2 : x^2-4x=5 please answer these . thanks 1 solutions Answer 280479 by jsmallt9(3296)   on 2011-01-16 00:07:57 (Show Source): You can put this solution on YOUR website!The solution provided by another tutor does not use the Quadratic Formula as the problem specifies and there is a typographics error in the first solution. question 1 : Dividing both sides by 2 will make the "a", "b" and "c" smaller. This will make the rest of the problem easier: The Quadratic Formula: which simplifies as follows: In long form this is: or or x = 14 or x = 9 question 2 : One side needs to be zero so we start by subtracting 5 from each side: The Quadratic Formula: which simplifies as follows: In long form this is: or or x = 5 or x = -1
 Mixture_Word_Problems/395146: 1. the sum of the digits f a two-digit number is 11. when the digits are reversed, the new number is increased by 20 which is twice the original number. find the original number. 2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number. pls help me!1 solutions Answer 280474 by jsmallt9(3296)   on 2011-01-15 23:49:37 (Show Source): You can put this solution on YOUR website!The solution to the first problem provided by another tutor is correct. The solution to the second problem has an error. 2. the units digit of a two-digit number is 3 less than its tens digit. if the number is divided by the sum of its digits, the quotient is 6 and the remainder is 8. find the number. The proper equations which express this are: u = t-3 We can eliminate the fractions in the second equation if we multiply both sides of the equation by (t+u): which simplifies as follows: 10t + u = 6*(t+u) + 8 10t + u = 6t + 6u + 8 Substituting t-3 for u we get: 10t + (t-3) = 6t + 6(t-3) + 8 which simplifies to: 11t - 3 = 6t + 6t - 18 + 8 11t - 3 = 12t - 10 Subtracting 11t from each side: -3 = t - 10 Adding 10 to each side: 7 = t This makes u = 7-3 or 4. The original number is 74.
 Exponential-and-logarithmic-functions/395078: Here is the problem I am working on: A bird species in danger of extinction has a population that is decreasing exponentially. Five years ago the population was at 1400 and today only 1000. Once the population drops below 100, the situation will be irreversible. When will this happen? I will use the model A=Aoe^kt I would appreciate any help! Thank You!1 solutions Answer 280469 by jsmallt9(3296)   on 2011-01-15 23:32:07 (Show Source): You can put this solution on YOUR website!First we use general exponential model equation and the initial information to find the specific exponential equation for this problem. In the general equation, the represents "the A when t = 0. The earliest value for A we have is from five years ago. So = 1400. This makes the other data, 1000 birds (A = 1000) five years later (t = 5) another point that should fit this equation. We now know 3 of the 4 unknowns in the general model. We can use the equation to find the 4th unknown, k: First we divide by 1400: which simplifies to Next we find the natural log of each side: Then we use a property of logarithms, , to move the exponent of the argument out in front. (It is this property that is the very reason we use logarithms on equations where the variable is in an exponent!) By definition, ln(e) = 1 so this becomes: Dividing by 5 we get: This is an exact expression for k. This makes the specific equation to model this problem: For practical use we should replace the k with its decimal approximation (which we can find using our calculator): We can now use this equation to find when the bird population drops below 100: We solve this for t, just like we solved the earlier equation for k: 39.3659175578729793 = t So the bird population will drop below 99 approximately 39.4 years after the t=0 time. Since the t=0 time was 5 years ago, the bird population will drop below 100 approximately 34.4 years from now.
 Rational-functions/395017: how do you solve 1 solutions Answer 280356 by jsmallt9(3296)   on 2011-01-15 14:24:04 (Show Source): You can put this solution on YOUR website!Another tutor's solution is mostly correct. But there is a mistake. Square both sides: (Subtracting 2x from each side (this is where the mistake was made): Factor: (x-2)(x-8) = 0 From the Zero Product Property we know that this (or any) product can be zero only if one (or more) of the factors is zero. So: x-2 = 0 or x-8 = 0 Solving these we get: x = 2 or x = 8 Whenever you square both sides of an equation extraneous solutions may be introduced. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. These extraneous solutions can occur even if no mistakes have been made! For this reason you must check your answers any time you square both sides of an equation. When checking, always us the original equation: Checking x = 2: Check failed! This is an extraneous solution and must be rejected. Checking x = 8: Check! So the only solution to your equation is x = 8.
 logarithm/394362: solve the equation to the nearest thousandth: 2^x+1=121 solutions Answer 280085 by jsmallt9(3296)   on 2011-01-14 13:25:14 (Show Source): You can put this solution on YOUR website!I'm guessing that the equation is: and not If this is correct, then please use parentheses on multiple term exponents. For example: 2^(x+1) = 12 Tutors are more likely to help if the problem is stated clearly. Solving equations where the variable is in an exponent usually involves logarithms. The base of the logarithm used does not make a significant difference. But if you want a decimal approximation (and you problem asks for a decimal), then you should choose a base of logarithm that your calculator "knows", like base 10 or base e (aka ln). I'm going to use base e logarithms. (If you choose base 10 instead the answer works out the same!) On the left side we can use a property of logarithms, , to move the exponent in the argument out in front. It is this very property that is the reason we use logarithms on problems like this. By moving the exponent, where the variable is, out in front the variable can now be solved for. Using this property of our equation we get: Now we can solve for x. Dividing by ln(2) we get: Subtracting 1 we get: This is an exact expression for the answer. For the decimal approximation we go to our calculators to find the two logarithms: x = 3.5849625007211562 - 1 which simplifies to: x = 2.5849625007211562 Rounded to the nearest thousandth we get: x = 2.585
 logarithm/394711: What is log 5.4 4x as a ratio of common logs1 solutions Answer 280079 by jsmallt9(3296)   on 2011-01-14 13:12:02 (Show Source): You can put this solution on YOUR website! A ratio of common logs? A ratio of logs should bring to mind the base conversion formula: . So all we have to do is use this formula to convert the base 5.4 log into a ratio of common (i.e. base 10) logs:
 Exponents/385295: Simplify: (-8)^-4/31 solutions Answer 279590 by jsmallt9(3296)   on 2011-01-12 23:02:14 (Show Source): You can put this solution on YOUR website! If you have trouble with negative and/or fractional exponents, I find it helpful if you factor the exponent in a certain way:If the exponent is negative, factor out a -1.If the exponent is a fraction and the numerator is not a 1, factor out the numerator. (You'll undetsand this better shortly.) Factoring your exponent this way we get: Looking at the exponent's factors we see:A -1 which tells us that we will find a reciprocal.A 4 which tells us that we will be raising to the 4th power.A 1/3 which tells us that we will be finding a cube root. And since multiplication is Commutative, we can do these three things in any order we choose! So we can shoose to do these things in the order that makes it easiest. (Note: What is easy for me is not necessarily easy for you. But don't worry. No matter what order you choose your answer will be the same as mine (as long as you do everything correctly) ) The order I choose is:A cube root. Normally this is not something that would be easy. But since a cube root first looks easy.Raise to the 4th power. The cube root in the first step will give us a fairly low number to raise to the fourth power.A reciprocal. This will introduce a fraction and fractions make things a little more difficult so why not put it off until the end? Here's the whole problem from the beginning to the end:
 Exponential-and-logarithmic-functions/393466: hello the question is 1/125=25^2x-1 FIND X:1 solutions Answer 279457 by jsmallt9(3296)   on 2011-01-12 16:03:35 (Show Source): You can put this solution on YOUR website!Assuming the equation is: (If this is correct, then please put parentheses around multiple term exponents like 2x-1.) When the variable in an equation is in an exponent, logarithms are often used to solve the problem. (And I will solve this way.) But if you can make both sides of the equation powers of the same number, then there is a much easier way to solve the equation. Since 125 and 25 are both powers of 5, this easier solution is possible for this equation. I'll start with the quick solution. First we rewrite each side as powers of 5. Since and since negative exponents are reciprocals, . So now we have: On the right side, the rule for raising a power to a power is to multiply the exponents. This gibes us: which simplifies to: The only way for two powers of 5 to be equal is for the exponents to be equal. So: -3 = 4x-2 Adding 2 we get: -1 = 4x Dividing by 4 we get: If we were not able to rewrite both sides of the equation as powers of the same number (or if we didn't notice that we could) then we use logarithms to solve equations like this. The base of logarithm is not really important. But if you want a decimal approximation for the answer, choose a base of logarithm that your calculator "knows", like base 10 or base e (aka ln). I'll use base 10 logarithms: Next we use a property of logarithms, , to move the exponent out in front as a coefficient. It is this very property that is the reason for using logarithms on equations like this. It allows us to move the exponent (where the variable is) out where we can solve for the variable. Using this property on the right side we get: We can now solve for x. Dividing both sides by log(25) we get: adding 1: Dividing by 2: Now we get out our calculators ans simplify the left side. You should get -0.25 (or some decimal extremely close to this). Since -0.25 = -1/4 this is the same answer we got the quick way.