Basic Logarithms
- Thread starter pangb
- Start date
Hello, pangb!
If you're just starting logs, this problem is awful!
I must assume that there is a typo.
We have: .\(\displaystyle \log_a(5\cdot 2) \:=\:1\)
. . . . . . . . . \(\displaystyle \log_a(10) \:=\:1\)
This means: . . . . \(\displaystyle a^1 \:=\:10\)
Therefore: . . . . . . \(\displaystyle a \:=\:10\)
If you're just starting logs, this problem is awful!
I must assume that there is a typo.
We have: .\(\displaystyle \log_a(5\cdot 2) \:=\:1\)
. . . . . . . . . \(\displaystyle \log_a(10) \:=\:1\)
This means: . . . . \(\displaystyle a^1 \:=\:10\)
Therefore: . . . . . . \(\displaystyle a \:=\:10\)
D
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Then there is "no solution" <<< That statement is incorrect - Mark has shown the solution below
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mmm4444bot
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The original equation does not seem too bad, if you use the change-of-base formula.
loga(5) = ln(5)/ln(a)
loga(2) = ln(2)/ln(a)
Substitute these expressions and clear the fraction to get:
ln(5)*ln(2) = ln(a)*ln(a)
Take the square root of both sides.
|ln(a)| = sqrt[ln(5)*ln(2)]
ln(a) = ±sqrt[ln(5)*ln(2)]
Switch to exponential form, to finish. :cool:
loga(5) = ln(5)/ln(a)
loga(2) = ln(2)/ln(a)
Substitute these expressions and clear the fraction to get:
ln(5)*ln(2) = ln(a)*ln(a)
Take the square root of both sides.
|ln(a)| = sqrt[ln(5)*ln(2)]
ln(a) = ±sqrt[ln(5)*ln(2)]
Switch to exponential form, to finish. :cool:
mmm4444bot
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The number e is a famous constant in mathematics. (Google "Euler's Number".)
As Jeff already said, when we use the number e as the base to a logarithm, we call the log a "natural logarithm".
In other words, loge(a) is the natural logarithm.
We almost always write the natural logarithm as ln(a).
So the name "ln" is just an abbreviation for "loge". You should be learning all of this very soon.
When we work with logarithms, sometimes it is handy to switch the base to e. In other words, we sometimes like to work with ln(x) instead of logb(x) -- particularly when the base b is not 10.
For example, what is log3.75(22) ?
That is, to what number does the base 3.75 need to be raised in order for the power to equal 22?
Our scientific calculator does not have a button for base-3.75 logarithms. Hence, we employ the change-of-base formula to convert the logarithm into the natural base because scientific calculators do have an LN button.
Here is the change-of-base formula: logb(x) = ln(x)/ln(b)
In other words, to convert some base-b logarithm to an equivalent expression that uses base-e instead, we take the natural log of the argument and divide that by the natural log of the base.
log3.75(22) = ln(22)/ln(3.75) = 2.3386 (rounded)
So now we know: the power 3.75^2.3386 equals 22 (rounded).
Another skill we need, when working with logarithms, is the ability to switch back and forth between what are called the "exponential form" and "logarithmic form".
Exponential form: b^n = x
Logarithmic form: logb(x) = n
When the base is Euler's Number (e), then these forms are:
e^n = x
ln(x) = n
In your exercise, we found that:
ln(a) = sqrt[ln(5)*ln(2)]
ln(a) = -sqrt[ln(5)*ln(2)]
To solve for a, we switch each of these statements to exponential form:
e^(sqrt[ln(5)*ln(2)]) = a
e^(-sqrt[ln(5)*ln(2)]) = a
We use our calculator to get decimal approximations for these two values of a.
I'm not sure how your instructor wanted you to approach this exercise, without first teaching the natural logarithm. Has your class ever used graphing calculators to zoom-in on a particular graph point, in order to estimate a solution?
As Jeff already said, when we use the number e as the base to a logarithm, we call the log a "natural logarithm".
In other words, loge(a) is the natural logarithm.
We almost always write the natural logarithm as ln(a).
So the name "ln" is just an abbreviation for "loge". You should be learning all of this very soon.
When we work with logarithms, sometimes it is handy to switch the base to e. In other words, we sometimes like to work with ln(x) instead of logb(x) -- particularly when the base b is not 10.
For example, what is log3.75(22) ?
That is, to what number does the base 3.75 need to be raised in order for the power to equal 22?
Our scientific calculator does not have a button for base-3.75 logarithms. Hence, we employ the change-of-base formula to convert the logarithm into the natural base because scientific calculators do have an LN button.
Here is the change-of-base formula: logb(x) = ln(x)/ln(b)
In other words, to convert some base-b logarithm to an equivalent expression that uses base-e instead, we take the natural log of the argument and divide that by the natural log of the base.
log3.75(22) = ln(22)/ln(3.75) = 2.3386 (rounded)
So now we know: the power 3.75^2.3386 equals 22 (rounded).
Another skill we need, when working with logarithms, is the ability to switch back and forth between what are called the "exponential form" and "logarithmic form".
Exponential form: b^n = x
Logarithmic form: logb(x) = n
When the base is Euler's Number (e), then these forms are:
e^n = x
ln(x) = n
In your exercise, we found that:
ln(a) = sqrt[ln(5)*ln(2)]
ln(a) = -sqrt[ln(5)*ln(2)]
To solve for a, we switch each of these statements to exponential form:
e^(sqrt[ln(5)*ln(2)]) = a
e^(-sqrt[ln(5)*ln(2)]) = a
We use our calculator to get decimal approximations for these two values of a.
I'm not sure how your instructor wanted you to approach this exercise, without first teaching the natural logarithm. Has your class ever used graphing calculators to zoom-in on a particular graph point, in order to estimate a solution?
mmm4444bot
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Oops! "base10" is a typographical error, yes?
I'm hoping that you meant to type "change base a to base e", instead.
