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# Recent problems solved by 'jsmallt9'

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 Polynomials-and-rational-expressions/278130: Can someone help me to solve this problem? 3x/x+2 + 6/x = 12/x^2+2x1 solutions Answer 202471 by jsmallt9(3296)   on 2010-03-07 19:26:11 (Show Source): You can put this solution on YOUR website! Equations are easier if there are no fractions, I think you'll agree. So we'll start by eliminating the fractions. The asiest way to eliminate the fractions in an equation is to multiply both sides by the Lowest Common Denominator (LCD). And to find the LCD we need to factor the denominators: The LCD is the product of all the different factors. So the LCD here is: (x+2)x This is what we will multiply both sides by: On the left side we need to use the Distributive Property: Now we can cancel: leaving: which simplifies as follows: Without the fractions, this is a very simple equation to solve. It is a quadratic equation so we want one side equal to zero. So subtract 12 from each side: Now we factor: 3x(x + 2) = 0 From the Zero Product Property we know that this product can be zero only if one of the factors is zero. So: 3x = 0 or x+2 = 0 Solving these we get: x = 0 or x = -2 With equations where the variable is in one or more denominators, it is important to check your answers. We must make sure no denominators are zero! Always check with the original equation. Checking x = 0: which simplifies to: As you can see, two of the denominators are zero. For this reason we must reject x = 0 as a solution. (If even only one denominator was zero we would still reject the solution.) Checking x = -2: which simplifies to: Again we get zero denominators! We must reject this solution, too. So there are no solutions to your equation!
 Polynomials-and-rational-expressions/278131: Find the zeros of the following polynomial. f(x)= 3x^3-55x^2=273x-851 solutions Answer 202467 by jsmallt9(3296)   on 2010-03-07 19:02:51 (Show Source): You can put this solution on YOUR website!Your "equation" has two equals sigms. Please correct and repost.
 Polynomials-and-rational-expressions/278133: Can someone please help me solve this problem? 2/5x+5 - 3/x^2-1 = 4/x-11 solutions Answer 202465 by jsmallt9(3296)   on 2010-03-07 19:00:44 (Show Source): You can put this solution on YOUR website! Equations are easier if there are no fractions, I think you'll agree. So we'll start by eliminating the fractions. The asiest way to eliminate the fractions in an equation is to multiply both sides by the Lowest Common Denominator (LCD). And to find the LCD we need to factor the denominators: The LCD is the product of all the different factors. So the LCD here is: (5)(x+1)(x-1) This is what we will multiply both sides by: On the left side we need to use the Distributive Property: Now we can cancel: leaving: (x-1)(2) - (5)3 = 5(x+1)4 which simplifies to: 2x - 2 - 15 = 20(x+1) 2x - 17 = 20x +20 Without the fractions, this is a very simple equation to solve. Subtract 2x from each side: -17 = 18x + 20 Subtract 20 from each side: -37 = 18x Divide both sides by 18: -37/18 = x With equations where the variable is in one or more denominators, it is important to check your answers. We must make sure no denominators are zero! A quick mental check and you can see that -37/18 does not make any of the denominators zero. So that is out solution.
 Polynomials-and-rational-expressions/278156: (8x^3-4x^2-6x-36)/(x-4) Thanks for your help.1 solutions Answer 202451 by jsmallt9(3296)   on 2010-03-07 18:39:26 (Show Source): You can put this solution on YOUR website! Since this fraction does not simplify. I'm guessing that the actual fraction is: Simplifying fractions involves canceling factors that are common to the nnumerator and denominator. To cancel factors we need factors. So we start by factoring. The numerator has a GCF of 2 which we can factor out. And the denominator is a difference of squares so it factors easily. We don't have common factors yet so we keep factoring. The second factor in the numeratordoesn't fit any of the factoring patternshas too many terms for trinomial factoringdoesn't appear to be factorable by grouping. So it seem that we need to factor by trial and error of the possible rational roots. The possible rational roots are the ratios, positive and negative, which can be formed using a factor of the constant term (at the end, 18 in your case) over a factor of the leading coefficient, 4. The factors of 18 are 1, 2, 3, 6, 9 and 18. The factors of 4 are 1, 2 and 4. So there are a lot of possible rational roots. But we're only interested in factors that match a factor in the denominator. So the only roots worth trying here are 2 and -2. I'm going to try 2 first. To check a rational root is probably easiest with synthetic division: ```2 | 4 -2 -3 -18 8 12 18 --------------- 4 -6 9 0 ``` The remainder is zero so 2 is a rational root and (x-2) is a factor. And, from the numbers in front of the remainder, the other factor is . So now our factored fraction is: We could try to continue to factor the numerator but we can see that the rational roots of do not include -2. So x+2 will not be a factor. Since any other factors do not help we won't bother factoring any more. Now we can cancel the common factor of x-2: which simplifies to:
 Polynomials-and-rational-expressions/277255: 2n to the second power+15n+7 I am learning factoring trinomials of the typer ax to the second power+bx+c1 solutions Answer 202400 by jsmallt9(3296)   on 2010-03-07 17:10:20 (Show Source): You can put this solution on YOUR website! So square the 2n: Now it is in the form ax^2 + bx + c
 Polynomials-and-rational-expressions/277475: the sum of the zeroes of the polynomial p(x) = 3(2x+7)^2(x-1)^2-(2x+7)(x-1)^3 is1 solutions Answer 202399 by jsmallt9(3296)   on 2010-03-07 17:06:35 (Show Source): You can put this solution on YOUR website! To find zeros we need to factor the polynomial. The straightforward way to factor this would be to simplify it first. But this is a lot of work. We would have to square 2x+7 and both square and cube x-1, add the like terms and then try to factor what we end up with. But you may notice that there are already common factors on each side of the subtraction in the middle. We can use this to our advantage and make the problem much easier. First I am going to rewrite the expression to make what I do clearer: The GCF of the expressions on each side of the subtraction are in red. I will now factor out this GCF from each side: Now I'll simply expression in the last factor: And we now have p(x) factored. And the zeros will be the values of x that make the factors zero: 2x+7 = 0 or x-1 = 0 or 5x+22 = 0 Solving these we get: x = -7/2 or x = 1 or x = -22/5 These are the zeros of p(x).
 Polynomials-and-rational-expressions/277466: if (2x^3+4x^2+2ax+b) is exactly divisible by (x^2-1),then the value of 'a' and 'b' respectively will be1 solutions Answer 202393 by jsmallt9(3296)   on 2010-03-07 16:46:35 (Show Source): You can put this solution on YOUR website!"Exactly divisible" means that there will be no remainder if you divide. So let's divide and see what "a" and "b" have to be in order for there to be no remainder: ``` 2x + 4 _______________________________ x^2-1 /2x^3 + 4x^2 + 2ax + b 2x^3 - 2x --------------------- 4x^2 + 2ax + 2x + b 4x^2 - 4 ------------------- 2ax + 2x + b + 4 ``` Since we're supposed to have no remainder, the expression at the bottom must work out to be zero. And the expression will be zero if the like terms combine to make zeros. So ```2ax + 2x = 0 and b + 4 = 0 2x(a + 1) = 0 b = -4 a = -1 ```
 Exponential-and-logarithmic-functions/277938: x^-1/7 x^7/61 solutions Answer 202385 by jsmallt9(3296)   on 2010-03-07 16:11:55 (Show Source): You can put this solution on YOUR website! When multiplying factors with exponents and the same base, like yours, the rule is to add the exponents. It does not matter what kinds of numbers the exponents are. just add them: As always, when you add fractions you must have the same denominator: which gives us:
 logarithm/277895: The graph of the equation y = 10^x lies entirely in Quadrants1 solutions Answer 202383 by jsmallt9(3296)   on 2010-03-07 15:52:19 (Show Source): You can put this solution on YOUR website!Each quadrant represents a certain combination of values for x and y: ``` x y Quadrant I Positive Positive Quadrant II Negative Positive Quadrant III Negative Negative Quadrant IV Positive Negative ``` So if we can figure out what values x can be and what values y can be, then we can figure out what what quadrants the graph will be in. In your equation x is the exponent (of 10). What kinds of numbers can exponents be? Answer: Exponents can be any number (positive, negative, zero, whole numbers, fractions, etc.). So x can be any number. In your equation the y is . What kinds of numbers can powers of 10 be? With some careful thought we should be able to determine that powers of 10 can never be zero or negative. Remember:A zero exponent of 10 results in 1, not zero: Negative exponents of 10 do not result in negative answers. For example, So we have found that x can be anything but y has to be positive. In which quadrants are positive y's found? The answer to this question is the answer to your problem.
 Exponential-and-logarithmic-functions/277816: Solve Equation 7^(x+1)=43 Answer 0.933 Help is appreciated. 1 solutions Answer 202220 by jsmallt9(3296)   on 2010-03-06 16:05:01 (Show Source): You can put this solution on YOUR website! Solving equations where the variable is in an exponent usually involves using logarithms. If you want an exact answer, then you can use a logarithm of any base. But the "best" base to choose the same as the base for which the variable is part of the exponent. In this case this would be base 7. But if you want a decimal approximation then you should choose a base your calculator "knows" (like base 10 or base e (ln)). I'll do the problem both ways. Using base 7 logarithms to get the simplest exact answer: Now we use a property of logarithms, , to move the exponent out in front. This property is precisely the reason we use logarithms on problems like this. It allows us to move the variable out of the exponent where we can "get at it". Using this property on the first logarithm we get: By definition, . (This is why we chose base 7 logarithms.) So the equation simplifies to: Now we just subtract 1 from each side: This is an exact expression for the solution to your equation. Using base 10 (or base e) logarithms (because we want a decimal approximation): Using the property to move the exponent out in front: Dividing both sides by log(7): Subtracting 1 from each side: Now we can get out our calculators andFind the two logarithms.Divide the two logarithmsSubtract 1 from the result of the division. This should result in something very close to the 0.933. P.S. If you find the exact answer then decide that you want a decimal, then you can take and use the base conversion formula, , to convert the base 7 logarithm into base 10 (or base e) logarithms. P.P.S. If you use base e instead of base 10 logarithms to find a decimal approximation, you end up with the same answer! (Try it an see.)
 logarithm/277821: Find the value of the expression log8 32 Ans:5/31 solutions Answer 202216 by jsmallt9(3296)   on 2010-03-06 15:44:27 (Show Source): You can put this solution on YOUR website! As it is, this logarithm cannot be found. No calculator "knows" base 8 logarithms and 32 is not a known power of 8. So what we need to do is transform this logarithm into an expression we can find. Fortunately there is a base conversion formula, , which we can use to change this base 8 logarithm into a logarithm we can find. So we need to choose a base to change the base 8 logarithm into. I'll choose two different bases. Base 2. Since 8 is a power of 2 (), using base 2 has some promise of making the problem easier. Using the conversion formula to convert the base 8 logarithm into an expression of base 2 logarithms we get: As stated earlier 8 is 2 cubed. So the logarithm in the denominator is 3. And, as it turns out 32 is also a power of 2! . So the logarithm in the numerator is 5. This makes our expression: 5/3 And we are finished (except perhaps changing the improper fraction into a mixed number). If 8 had not been a power of 2 or if we didn't recognize that 8 (or 32) was a power of 2, then we would need to cahnge the base to one our calculator "knows" (like base 10 or base e (ln)). Converting your expression into base 10 logarithms we get: And now we would use our calculators to find the two logarithms and them divide. The first way is better than this way. For one, calculators use decimal approximations for most logarithms. Secondly, the decimal answer you get from a calculator may by hard to recognize as equal to the fraction 5/3.
 logarithm/277571: [inside brackets = Base] (log[5](3x+10))-(3log[5](4))=2 Solve the equation. Thanks.1 solutions Answer 202148 by jsmallt9(3296)   on 2010-03-06 06:54:50 (Show Source): You can put this solution on YOUR website! We want the equation in the form: log(expression) = other-expression So somehow we need to combine the two logarithms into one. These two logarithms are not like terms so we cannot subtract them. But there is a property of logarithms, , which can be used to combine two logarithms if all of the following are true:there is a "-" between themthe bases of the logarithms are the samethe coefficients of the logarithms are 1's Your logarithms meet the first two but not the last. So now our goal is to get rid of the 3 in front of the second log. And fortunately there is another property of logarithms, , which can be used to move a coefficient of a logarithm into its argument as an exponent. Using this on your second log we get: which simplifies to: These are still not like terms so we still cannot subtract them. But we can now use the other property to combine them: We now have the desired form. Once we have this form the next step is to rewrite the equation in exponential form: which simplifies to: Now the variable is out of the argument where we can "get at it". Solving this for x we start by multiplying both sides by 64 to get rid of the fraction: 3x + 10 = 1600 Subtracting 10 from each side: 3x = 1590 Dividing by 3 we get: x = 530 When solving logarithmic equations, it is important (not just a good idea) to check your answers. Even if we've done everything correct so far, we need to make sure that each answer makes the argument of any logarithms positive. Always use the original equation to check: Checking x = 530: which simplifies to Both arguments are positive so it looks good. (You're welcome to finish the check on your own.) The solution to your equation, then, is x = 530}}}
 logarithm/277569: 2logx - log5 = 41 solutions Answer 202147 by jsmallt9(3296)   on 2010-03-06 06:41:04 (Show Source): You can put this solution on YOUR website! We want the equation in the form: log(expression) = other-expression So somehow we need to combine the two logarithms into one. These two logarithms are not like terms so we cannot subtract them. But there is a property of logarithms, , which can be used to combine two logarithms if al of the following are true:there is a "-" between themthe bases of the logarithms are the samethe coefficients of the logarithms are 1's Your logarithms meet the first two but not the last. So now our goal is to get rid of the 2 in front of the first log. And fortunately there is another property of logarithms, , which can be used to move a coefficient of a logarithm into its argument as an exponent. Using this on your first log we get: These are still not like terms so we still cannot subtract them. But we can now use the other property to combine them: We now have the desired form. Once we have this form the next step is to rewrite the equation in exponential form: which simplifies to: Now the variable is out of the argument where we can "get at it". Solving this for x we start by multiplying both sides by 5 to get rid of the fraction: So or Simplifying these square roots we get: or When solving logarithmic equations, it is important (not just a good idea) to check your answers. Even if we've done everything correct so far, we need to make sure that each answer makes the argument of any logarithms positive. Always use the original equation to check: Checking Both arguments are positive so it looks good. (You're welcome to finish the check on your own.) Checking As you can see the argument to the first logarithm is negative. We cannot allow this to occur. So we reject this solution. The only solution to your equation, then, is
 logarithm/277539: find the inverse of the function y=log5(x^2) the five is small1 solutions Answer 202070 by jsmallt9(3296)   on 2010-03-05 13:56:24 (Show Source): You can put this solution on YOUR website! To find an inverse:Swap the x's and y's. Wherever you see an x, write y and wherever you see y, write x. This creates the inverse equation.If possible solve the new equation for y Let's see how this works on your equation. 1. Swap the x's and y's: 2. Solve for y, if possible. To solve our equation for y we start by rewriting the equation in exponential form. (This gets y out the of the argument of the logarithm.) At this point we get stuck. We cannot solve this for y. (At least not for a single-valued expression for y.) So the inverse of your original function is not a function. The inverse can be expressed with any of the following: ( or )