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# how to graph y=2log3(x-4)+5

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1879
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how to graph y=2log3(x-4)+5

Jun 5, 2015

#2
+97547
+8

$$y=2log_3(x-4)+5$$

You can BUILD this graph,

you need to have some idea of what  $$y=log_3x$$     looks like.

You should know the basic shape of a log function and know that it will go through the point (1,0)

also since it is  the same as  $$x=3^y$$    other points will be (3,1),  (9,2), (1/3,-1)  etc

Once you have drawn   $$y=log_3x$$    you can start manipulating it.

To get   $$y=log_3(x-4)$$     you just move the whole graph 4 places to the right  (positive direction)

then

To get   $$y=2log_3(x-4)$$     all the y values are doubled so the range is doubled.

then

to get    $$y=2log_3(x-4)+5$$    just move the whole graph UP 5 places.

I have done these steps here

https://www.desmos.com/calculator/0cslw2km9u

Initially only the   $$y=log_3x$$    graph is showing.

The equations are on the left.  There are circles next to them.  click on the circles one at a time - top to bottom - and you will see your graph get Built.

Jun 6, 2015

#1
+27475
+8

If you are doing it by hand choose several values of x (greater than 4) and calculate the corresponding values of y, then plot the pairs of values on graph paper.

.

Jun 6, 2015
#2
+97547
+8

$$y=2log_3(x-4)+5$$

You can BUILD this graph,

you need to have some idea of what  $$y=log_3x$$     looks like.

You should know the basic shape of a log function and know that it will go through the point (1,0)

also since it is  the same as  $$x=3^y$$    other points will be (3,1),  (9,2), (1/3,-1)  etc

Once you have drawn   $$y=log_3x$$    you can start manipulating it.

To get   $$y=log_3(x-4)$$     you just move the whole graph 4 places to the right  (positive direction)

then

To get   $$y=2log_3(x-4)$$     all the y values are doubled so the range is doubled.

then

to get    $$y=2log_3(x-4)+5$$    just move the whole graph UP 5 places.

I have done these steps here

https://www.desmos.com/calculator/0cslw2km9u

Initially only the   $$y=log_3x$$    graph is showing.

The equations are on the left.  There are circles next to them.  click on the circles one at a time - top to bottom - and you will see your graph get Built.

Melody Jun 6, 2015
#3
+4682
+5

log...

Hmm, i always see that word.

My brother said it's very hard to understand...

Jun 6, 2015
#4
+97547
+5

Hi again MathsGod,

Basically a log is a power.

For instance

$$\\2^3=8\\\\ so\\\\ log_28=3$$

See how the answer to the log is the power   (or exponent)

BUT

Your brother is correct, they are difficult to understand and you are nowhere near ready to deal with them yet!

Jun 6, 2015
#5
+4682
+5

So it likes the inverse?

2^3=8

log_2^8=3

Jun 6, 2015
#6
+97547
+5

I am not sure what the right word is

but yes they can be used to cancel each other out.

Like + and -

Or

× and ÷

I'm not explaining this very well. :/

Jun 6, 2015
#7
+4682
+5

Haha, don't worry i understand that very well.

It's like you do the opposite to get the inverse.

1+2=3

3-2=1

2*3=6

6/3=2

3^2=9

sqrt(9)=3

Now that I'm doing Square root, i have a question about it, when doing e.g. 5^5

How do you do the inverse in sqrt on the calculator.

4^2=16

sqrt(16)=4

But...

2^5=32

...(5)sqrt(32)???

Jun 6, 2015
#8
+97547
+5

Hi MathsGod1

Yes, that is all correct!

Umm let me see

$$\\4^2=16\\ \sqrt{16}=4\qquad That is certainly true\\ But a square root is really a power of 1/2 so\\ \sqrt{16}=16^{1/2}=4\\\\$$

I just had a thought, it will help you with equations too.  I think inverse is the right word

6+5-5=6                  plus 5 is the inverse of minus 5

$$3*6\div 6 = 3$$        divide by 6 is the inverse of multiply by 6

$$(\sqrt{7})^2=7$$             square is the inverse of square root

$$\sqrt{17^2}=17$$               square root is the inverse of square

$$\\\sqrt[5]{12} =12^{1/5}\;\;and\;\;\\\\ (12^{1/5})^5=12^{5/5}=12\qquad The power of 5 is the inverse of the fifth root$$

$$log_{10}10^7=7$$                   the log is the inverse of base number.  Again not wording this properly.

Jun 7, 2015
#9
+4682
+5

Cool!

But what is:

x^1/2

Jun 7, 2015
#10
+97547
0

$$\\x^{1/2}=\sqrt{x}\\\\ x^{1/3}=\sqrt[3]{x}\\\\ x^{1/4}=\sqrt[4]{x}\\\\ x^{1/5}=\sqrt[5]{x}\\\\ x^{1/6}=\sqrt[6]{x}\\\\\\ 81^{1/4}=\sqrt[4]{81}=3\\\\ This is because  3*3*3*3 = 3^4 =81$$

.
Jun 7, 2015