SOLUTION: Show work. Good morning, Please I need help with these logarithmic problems.I am a seventh grader and I have no idea what I am doing. I tried some but these are the ones I having

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Show work. Good morning, Please I need help with these logarithmic problems.I am a seventh grader and I have no idea what I am doing. I tried some but these are the ones I having       Log On


   

Question 593988: Show work.
Good morning, Please I need help with these logarithmic problems.I am a seventh grader and I have no idea what I am doing. I tried some but these are the ones I having problems with. I am totally lost and I have to turn this in tomorrow. I pray that someone can help me today with these problems. I have been trying to figure it out all weekend. Thanks alot and my God bless you.
Convert to a logarithmic equation.
1.) e^5=p
Write the exponential equation
2.)log base2(k)=L
3. y=(1/7)^x: The graph of y(1/7)^x is translated six units to right reflected in x-axis and the translated two units upward . Write the equation of the graph in its final position.
4.) f(x)=-1/2log(x+1)Sketch the graph of functin, state domain and range of the function.
5.f(x)=IN(x+3)Find f^-1(x)
6.xarrowinfinity under(lim 8^2x)Use a graph to fin the following limit
7. logbase3(1/9).Evaluate the expression
8. f(x)=(1/4)^4. Cive the function find f(3)
9. Find the logarithm. logbase256^(4)
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, this is far too many questions for a single post. This is probably why it has taken so long for anyone to reply. For this reason I am going to answer some of the questions, give hints on others and maybe ignore the rest.

#1 & #2. In general
x%5Ey+=+z is equivalent to log%28x%2C+%28z%29%29+=+y

#3. A right shift of any function is effected by replacing the x's with (x-n) where n is the amount of the shift. (For left shifts, use a negative "n".)

So the graph of y+=+%281%2F7%29%5Ex shifted to the right by 6 would have an equation of y+=+%281%2F7%29%5E%28x-6%29

A reflection of a graph in the x-axis is effected by replacing the y with -y (and then solving for y). (To reflect in the y-axis replace x with -x.) So the graph of y+=+%281%2F7%29%5E%28x-6%29 reflected in the x-axis would have an equation of -y+=+%281%2F7%29%5E%28x-6%29. Solving for y (by multiplying each side by -1) we get:
y+=+-%281%2F7%29%5E%28x-6%29

A vertical shift upward is effected by replacing y with (y-n) where "n" is the amount of the shift. (Use a negative n for a downward shift.) So the graph of y+=+-%281%2F7%29%5E%28x-6%29 shifted upward two units would have an equation of y+-+2+=+-%281%2F7%29%5E%28x-6%29. Solving for y (by adding 2 to each side) we get:
y+=+-%281%2F7%29%5E%28x-6%29%2B2

#4. In general, logarithms can be any real number. And a real number times -1/2 is still a real number. So the range is all real numbers. For the domain of a logarithmic function you must restrict the values of x so that argument of the logarithm is always positive. So to find the domain, write an inequality that says that the argument is positive and solve that inequality. The solution will be the domain. For your problem you would write:
x+1 > 0
Whatever the solution is is the domain.

#5. A general procedure for finding inverses:
  1. If function notation is being used, like f(x), replace the function notation with "y".
  2. Swap the x's' and y's. (Just rewrite the equation using x's where the y's were and vice versa.) This is not an algebraic process. You are changing the equation into its inverse when you do this. (The inverse you now have, however, is not yet in its proper form. The rest of the procedure is to transform the equation into its proper form.)
  3. Solve the new equation for y. NOTE: If you cannot solve for y without using +-, then the inverse is not a function. (Hint: If your equation has logarithms, then you will need to rewrite it in exponential form to solve for y.)
  4. If the inverse is a function (See step #3) and if the problem started with functional notation, replace the y with the functional notation for the inverse, like f%5E%28-1%29%28x%29

#6. The limit does not exist for the expression you posted. I suspect that you left something out or mistyped something. If you have a graphing calculator and know how to use the graphing function, enter the correct equation look to see of the y coordinated of points on the far right (if the limit is to negative infinity look to the far left) get closer and closer to some value. If they do then that value is the limit.

#7. See #9

#8. There is no x in f(x). If this is correct then the answer is simply %281%2F4%29%5E4+=+1%2F256. If there is an x, replace it with a 3 and simplify the expression.

#9. Logarithms can be calculated "by hand" if
  • Case 1: The argument is a known power of the base. In this case the answer is simply that power. For example, log%282%2C+%288%29%29+=+3 since 8 (the argument) is 2 (the base) to the 3rd power.
  • Case 2: The base is a known power of the argument. In this case use the change of base formula, log%28a%2C+%28p%29%29+=+log%28b%2C+%28p%29%29%2Flog%28b%2C+%28a%29%29, to convert the base to the value of the argument. For example, log%288%2C+%282%29%29. Change the base to 2 (the argument):
    log%288%2C+%282%29%29+=+log%282%2C+%282%29%29%2Flog%282%2C+%288%29%29
    Both of the logs on the right are "case 1" logs (above). So
    log%288%2C+%282%29%29+=+log%282%2C+%282%29%29%2Flog%282%2C+%288%29%29+=+1%2F3
  • Case 3: Both the argument and the base are known powers of some third number. In this case use the change of base formula to change the base to that third number. For example. log%2832%2C+%2864%29%29. 32 and 64 are not well-known powers of each other. But they are both powers of 2. So we use the change of base formula to change to base 2 logs:
    log%2832%2C+%2864%29%29+=+log%282%2C+%2864%29%29%2Flog%282%2C+%2832%29%29
    Since 2%5E6+=+64 and 2%5E5+=+32 the logs on the right become 6 and 5 respectively:
    log%2832%2C+%2864%29%29+=+log%282%2C+%2864%29%29%2Flog%282%2C+%2832%29%29+=+6%2F5

For log%28256%2C+%284%29%29 you will have to use the change of base formula (unless you are clever enough to figure out what power of 256 is 4). For #7 you might be able to figure out what power of 3 results in 1/9. If not, you will have to use the change of base on this one, too.