I can't make sense of this log property explanation?

AI Thread Summary
The discussion revolves around two main mathematical queries: the relationship between logarithms and graphing a quadratic function. The first part clarifies that log(base A) of B equals 1/log(base B) of A through algebraic manipulation, emphasizing the definition of logarithms. The second part addresses confusion in graphing the equation y = x^2 + x + 1 for x ≤ 1, noting that the provided points do not align with the expected graph. It is highlighted that the equation is piecewise defined, with different expressions for x values less than or equal to 1 and greater than 1. Overall, the discussion seeks to resolve misunderstandings in both logarithmic properties and graphing techniques.
roger12
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Homework Statement



I decided to cram these two unrelated question into one post, because they are too small and I don't want to crowd the forum with my many little bitty questions.


1. log(base A) of B= 1/[log(base B) of A]

because: if log(base B) of A=C, then B^C=A and so B=A^1/C

Hence, log(base A) of B= 1/C= 1/[log(base B) of A]

2. plot the graph of:

x^2+x+1: x<1, x=1 for [-3, 4] with intervals of 0.5

It's a part of a bigger graph.


Homework Equations





The Attempt at a Solution



1. I don't see how introducing C as equal to a different base and then carrying out a bunch of algebraic manipulations prove anything. Am I missing something here? Thank You.

2. x,y pairs (-3, 7), (-2, 3), (-1,1) I chose the whole numbers for x, because they are more convenient to me. If you look at this part of the graph in my book it goes through different points. Can, you please tell me where I went wrong? Thanks.
 
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roger12 said:
1. log(base A) of B= 1/[log(base B) of A]

because: if log(base B) of A=C, then B^C=A...
Remember the for logarithms,
logn x = y iff ny = x

So, using this, if
logB A = C, then BC = A.

... and so B=A^1/C
From
BC = A,
raise both sides to the exponent of 1/C, so
(BC)1/C = (A)1/C,
or B = A1/C.

... Hence, log(base A) of B= 1/C= 1/[log(base B) of A]

Using that definition of logarithms I gave earlier, Since
B = A1/C,
logAB = 1/C.

Substitute logB A = C into the fraction, so
logAB = 1/(logB A)
 
roger12 said:

Homework Statement



2. plot the graph of:

x^2+x+1: x<1, x=1 for [-3, 4] with intervals of 0.5

It's a part of a bigger graph.

The Attempt at a Solution


2. x,y pairs (-3, 7), (-2, 3), (-1,1) I chose the whole numbers for x, because they are more convenient to me. If you look at this part of the graph in my book it goes through different points. Can, you please tell me where I went wrong? Thanks.

Apparently you are graphing the equation y = x2 + x + 1, although what you showed is not an equation.

What does this part (in red) mean?
x^2+x+1: x<1, x=1[/color] for [-3, 4] with intervals of 0.5

The points you show, (-3, 7), (-2, 3), (-1,1), are on the graph of y = x2 + x + 1. Does the graph in your book come from this equation?
 
eumyang said:
Using that definition of logarithms I gave earlier, Since
B = A1/C,
logAB = 1/C.

Substitute logB A = C into the fraction, so
logAB = 1/(logB A)

Awesome! Why I forgot "If A=B^C, then C=log(base B)A" is beyond me.
 
Mark44 said:
Apparently you are graphing the equation y = x2 + x + 1, although what you showed is not an equation.

What does this part (in red) mean?


The points you show, (-3, 7), (-2, 3), (-1,1), are on the graph of y = x2 + x + 1. Does the graph in your book come from this equation?

The parts in red mean x is less than or equal to 1.

And it IS an equation :

y=x^2+x+1: x< or =1

and

y= 3-x : x>1

for [-3, 4]

y is defined differently for different values of x. But both expressions are the parts of the same equation. The part of the graph defined by y= 3-x : x>1 for [-3, 4] looks right, but the other part defined by y=x^2+x+1: x< or =1 for [-3, 4] doesn't sit on the points made up of ordered pairs above.
 
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