Questions on Algebra: Exponent and logarithm as functions of power answered by real tutors!

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how to solve
299^99
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There's nothing to solve. Equations are solved, this is not one.
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The result of the calculation is a big number.
It's 1.23437649...E245, 240+ digits

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I would say the function is first.
Then, I would substitute the coordinates of the points given and look at what I get for ideas.


Once you look at those two equations, you may think of a good way to solve it, that would not give you much trouble. Even if you do not choose the easiest or fastest way to the solution, do not panic. Try to stay calm and confident.
ONE OPTION
You could decide on dividing the second equation by the first one
--> or -->
Once you get , you can substitute in to find .
--> or -->
So the function is
ANOTHER OPTION
You could decide on solving for in the first equation and substituting that into the second one.
For example, you found that and substituted to get

That may look scary, but can be simplified
--> --> --> --> -->
amd you can substitute that into to find , as shown above.
This option may look more complicated, but still leads to the solution.
AND YET ANOTHER OPTION
You could decide on solving for in the first equation and substituting that into the second one.
substituted into
--> --> --> --> --> --> --> -->
and you can substitute that into to find :
--> or -->
COMMENT
Staying calm is difficult, but necessary. Being confident is even harder. Having seriously practiced algebra in previous years will help give you confidence, and make calculation mistakes less likely. If you did not take algebra that seriously before, I can tell you that I've been there, and that there are plenty of people in the same boat. Several times, I found out the hard way that math, like languages and music, gets much harder the longer that you do not practice it seriously and regularly. Fortunately, at various points, I stumbled into very studious study buddies that made me practice math a lot more and be (temporarily) a little more responsible.

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show that log8 1000=log2 10 algebraically
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log(1000)/log(8) = log(10)/log(2)
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log(1000)/log(10) = log(8)/log(2)
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log10(1000) = log2(8)
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3 = 3
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Cheers,
Stan H.
====================

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rewrite equation in logarithmic form
64^(1/2) = 8
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base = 64
exponent = log = (1/2)
result = 8
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Log Form:
log64(8) = 1/2
===================
Cheers,
Stan H.
===================

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Use the fact that log_b (x) = y can be written b^y = x.

u^v = (15/16)

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Square both sides
assuming n-2 is the exponent. Use parentheses.
2n*log(2) = (n-2)*log(3) = nlog(3) - log(9)
2n*log(2) - n*log(3) = -log(9)
n*(log(3) - 2log(2)) = log(9)
n = log(9)/log(3/4)
n =~ -7.6377

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a.
If , then
add 1 to the both sides of the equation
divide by 2

use scientific calculator
b.
If , then

d.

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is NOT the antilogarithm. is the base of the natural logs, that is:

If one half-life has elapsed, then .

Since the half-life of the substance in question is 32 years, then



Take the natural log of both sides:



Use



to write



Use "the difference of the logs is the log of the quotient",



to write



Use



to write



then use



to write



then



John

My calculator said it, I believe it, that settles it

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Let me start by saying, I don't know if the problem is
or I will solve it both ways!

Solving the SECOND statement above:


Divide both sides by 8:


Raise both sides as a power of e:



Multiply both sides by 2:


Solving the FIRST statement above:
8 ln(1/2x) = 72

Divide both sides by 8:
ln(1/2x) = 9

Raise both sides as a power of e:



Cross multiply to get this:




For a complete explanation of this and other topics in LOGARITHMS, please see my own website. The best way to find it is to use the easy-to-remember and easy-to-spell link www.mathinlivingcolor.com. At the bottom of the page there is a single link that takes you to my Homepage. On my Homepage, look for the link "Basic, Intermediate, and College Algebra: One Step at a Time." Choose "College Algebra", and look in "Chapter 4" for a complete explanation of logarithms. Here you will find my own non-traditional explanation that my own students used to find much easier to understand than most of the other textbooks being used. I also have lots of examples, exercises, and answers. In addition, most of the hardest problems are solved in color on the MATH IN LIVING COLOR pages that go with this topic. There is TWO VIDEOS of me teaching LOGARITHMS in class before I retired a few years ago. To find the videos, look on my Homepage for "Rapalje Videos in Living Color," choose "College Algebra" and select the "Logarithms" videos. It's ALL FREE on my website!!!

To contact me please use my Email address at rapaljer@seminolestate.edu.

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus

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Use

The sum of the logs is the log of the product, the difference of the logs is the log of the quotient, and the log of something to a power is the power times the log:







to write:



Then substitute the values you were given plus the fact that to write:



And do the arithmetic.

John

My calculator said it, I believe it, that settles it

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solve for x:
log2(x+2) + log2(x+6)=5
log2[(x+2)(x+6)]=5
convert to exponential form: base(2) raised to log of number(5)=number(x+2)(x+6)
2^5=x^2+8x+12
32=x^2+8x+12
x^2+8x-20=0
(x+10)(x-2)=0
x=-10(reject, x>0)
or
x=2

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I presume you mean



Use



to write



Then use



to write



John

My calculator said it, I believe it, that settles it

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r=0.4855886

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Use the formula v= 4/3 pi r^3
1. If the radius of a ball is twice that of another, how many times as big is its volume?
8 times.


The volume is related to r^3. (n*r)^3 = n^3*r^3
Volume is a function of the cube of the radius.
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2. If the volume is twice that if a blue ball, how many times as big is its radius?
Volume is a function of the cube of the radius.
I don't know about the color, but the radius is times.

                  x2[log(x)]² = 10x3

Take logs of both sides:

             log{x2[log(x)]²} = log(10x3)

Use rules of logarithms:

        2[log(x)]²·log(x) = log(10) + log(x³)

               2[log(x)]³ = 1 + 3·log(x)

2[log(x)]³ - 3·log(x) - 1 = 0

let u = log(x)

             2u³ - 3u - 1 = 0

Possible rational solutions for u are ±1, ±

Try 1:

1|2  0 -3 -1
 |   2  2 -1
  2  2 -1 -2 

No that isn't a solution, since it did not give a remainder of 0.

Try -1:

-1|2  0 -3 -1
  |  -2  2  1
   2 -2 -1  0 

Yes -1 is a solution, so we have factored

             2u³ - 2u - 1 = 0
as

     (u + 1)(2u² - 2u - 1) = 0

   u + 1 = 0    2u² - 2u - 1 = 0
       u = -1             u = 
                          u = 
                          u = 
                          u = 
                          u = 
                          u = 
                          u = 
                          
The log equation u = log(x) is equivalent to the exponential
equation x = 10u

So we have three solutions:

x = 10-1, x = , x = 

In decimal approximations they are

x = 0.1,  x = 23.22872667,  x = 0.4305014278

Edwin