First, let's figure out what the base needs to be. What do you think? It looks like the b in the definition correlates with 2 in our problem - so our base is going to be 2.
Next, let's figure out the exponent. This is very key, again remember that logs are another way to write exponents. This means the log is set equal to the exponent, so in this problem that means that the exponent has to be 5.
That leaves 32 to be what the exponential expression is set equal to.
Putting all of this into the log definition we get:
First, let's figure out what the base needs to be. What do you think? It looks like the b in the definition correlates with b in our problem - so our base is going to be b.
Next, let's figure out the exponent. This is very key, again remember that logs are another way to write exponents. This means the log is set equal to the exponent, so in this problem that means that the exponent has to be 4.
That leaves 81 to be what the exponential expression is set equal to.
Putting all of this into the log definition we get:
First, let's figure out what the base needs to be. What do you think? It looks like the b in the definition correlates with 4 in our problem - so our base is going to be 4.
Next, let's figure out the exponent. In this direction it is easy to note what the exponent is because we are more used to it written in this form, but when we write it in the log form we have to be careful to place it correctly. Looks like the exponent is -3, don't you agree?
The value that the exponential expression is set equal to is what goes inside the log function. In this problem that is 1/64.
Let's see what we get when we put this in log form:
Rewriting the original problem using exponents we get:
First, let's figure out what the base needs to be. What do you think? It looks like the b in the definition correlates with 9 in our problem - so our base is going to be 9.
Next, let's figure out the exponent. In this direction it is easy to note what the exponent is because we are more used to it written in this form, but when we write it in the log form we have to be careful to place it correctly. Looks like the exponent is 1/2, don't you agree?
The value that the exponential expression is set equal to is what goes inside the log function. In this problem that is x.
Let's see what we get when we put this in log form:
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 3 to get 81
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 3 to get 1
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 11 to get 11
*Setting the log = to x
*Rewriting in exponential form
*x is the exponent
we need on 7 to get square root of 7
Looks like the base is 4, the exponent is y,
and the log will be set = to x:
The first two columns just show what values we are going to plug in for y.
The last three columns show the corresponding values for x and y for the given function.
x
y
y
(x, y)
-2
-2
(1/16, -2)
-1
-1
(1/4, -1)
0
0
(1, 0)
1
1
(4, 1)
2
2
(16, 2)
Next, we need to write in exponential form, just like we practiced in examples 1 and 2 on the lesson page.
Looks like the base is 4, the exponent is y - 1, and the log will be set = to x:
The first two columns just show what values we are going to plug in for y.
The last three columns show the corresponding values for x and y for the given function.
x
y
y
(x, y)
-2
-2
(1/64, -2)
-1
-1
(1/16, -1)
0
0
(1/4, 0)
1
1
(1, 1)
2
2
(4, 2)
Since x is part of the inside of the log
on this problem we need to find a value of x,
such that the inside of the log, 2 - x, is
positive.
*Domain of this function
That means that if we put in any value of x that is less than 2, we will end up with a positive value inside our log.
I'm going to use the first
inverse property shown on the lesson page:
Last revised on March 17, 2011 by Kim Seward.
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