I went online to this website: http://www.mathsisfun.com/algebra/expone…
where it says "working together " and "Properties of Logarithms"
It says that a^x=y and log a (y)=x
(is that what that chart says?)
It also says they are inverse. when I graphed a^x=y and log a (y)=x...( sorry i don't know how to make the a look ans smaller :( ) they were the same. Aren't inverse function, suppose to be a reflection of each other over the line y=x?
On the website, the part that is titled "Properties of Logarithms"
Why is this true?
Can anyone explain clearly? My teacher doesn't want us just to say that when u have multiply exponents you add the exponents.
where it says "working together " and "Properties of Logarithms"
It says that a^x=y and log a (y)=x
(is that what that chart says?)
It also says they are inverse. when I graphed a^x=y and log a (y)=x...( sorry i don't know how to make the a look ans smaller :( ) they were the same. Aren't inverse function, suppose to be a reflection of each other over the line y=x?
On the website, the part that is titled "Properties of Logarithms"
Why is this true?
Can anyone explain clearly? My teacher doesn't want us just to say that when u have multiply exponents you add the exponents.
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When you plot log a (y) = x, make sure that you are plotting log y = x*log a, since base 10 is the default setting. meaning that log (a number) = log 10 (a number), 10 is the base.
Also, a^x=y and log a (y)=x are the same function. Where as a^x=y and log a (x)=y are inverses of each other
Properties of Logarithms:
We want to show that: log a ( m × n ) = log a (m) + log a (n).
Okay, so lets k = log a (m) + log a (n)
So, k = log a (m) + log a (n). I am little bit too lazy to type a, so lets say a = 10, so that log a (m) = log m. But the derivation is general, so a could be anything greater than 0.
okay, back to our discussion, k = log (m) + log (n), so 10^k = 10^ (log (m) + log (n)). since log (m) has base 10, and 10^(log (m)) = m. Also, C^(E+D) = C^E*C^D
So, 10^k = 10^[log (m))] * 10^[ log (n)] => 10^k = m*n => k = log(m*n).
Since k = log (m) + log (n), and k = log(m*n). So, log(m*n) = log (m) + log (n).
Like I told you before, the result is general, so we do not have to stick with base 10. So, log a ( m × n ) = log a (m) + log a (n) is derived.
Also, a^x=y and log a (y)=x are the same function. Where as a^x=y and log a (x)=y are inverses of each other
Properties of Logarithms:
We want to show that: log a ( m × n ) = log a (m) + log a (n).
Okay, so lets k = log a (m) + log a (n)
So, k = log a (m) + log a (n). I am little bit too lazy to type a, so lets say a = 10, so that log a (m) = log m. But the derivation is general, so a could be anything greater than 0.
okay, back to our discussion, k = log (m) + log (n), so 10^k = 10^ (log (m) + log (n)). since log (m) has base 10, and 10^(log (m)) = m. Also, C^(E+D) = C^E*C^D
So, 10^k = 10^[log (m))] * 10^[ log (n)] => 10^k = m*n => k = log(m*n).
Since k = log (m) + log (n), and k = log(m*n). So, log(m*n) = log (m) + log (n).
Like I told you before, the result is general, so we do not have to stick with base 10. So, log a ( m × n ) = log a (m) + log a (n) is derived.
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a^x=y is the SAME equation as log(base a)(y)=x
take log(base a) of both sides to see that
log(base a) of a^x is asking what power do I have to raise a to in order to get a^x? of course, the anser is x.
to get the inverse function, swap the x and the y after you have solved for x
y = log(base a) (x) would be the inverse
take log(base a) of both sides to see that
log(base a) of a^x is asking what power do I have to raise a to in order to get a^x? of course, the anser is x.
to get the inverse function, swap the x and the y after you have solved for x
y = log(base a) (x) would be the inverse