Need help understanding logarithmic properties

[daigo]

Okay, so in class I learned that:

10^log(4x) = 4x

But I don't understand why.

I get that a log without a subscript is considered base 10, so:

10^y = 4x

Is the way to understand "log(4x)", right?

What if the problem was a different base? Would the "10" coefficient and 'log' in the original problem disappear also? If not, what coefficient would it have to be to match the base log in order for those to disappear?

 

[SteveL27]

Okay, so in class I learned that:

10^log(4x) = 4x

But I don't understand why.

Well, log(4x) is the power that you raise 10 to in order to get 4x.

And then you do it ... you raise 10 to that power ... so that 10^log(4x) = 4x.

In other words, log(4x) is some quantity that IF you raised 10 to that power, you'd get 4x. And then you raise 10 to that power.

A more sophisticated way of understanding this is that taking a log (with respect to some base) and raising that base to a power, are operations that are inverse to one another.

Keep doing problems, this will become clear with practice.

What if the problem was a different base? Would the "10" coefficient and 'log' in the original problem disappear also? If not, what coefficient would it have to be to match the base log in order for those to disappear?

It has to be the same base. So 3^(log_3(x)) = x

where log_3 is the base-3 logarithm.

Because log_3(x) is the power you'd have to raise 3 to in order to get x; and then you raise 3 to that power, so you get x.

 

[daigo]

It has to be the same base. So 3^(log_3(x)) = x

where log_3 is the base-3 logarithm.

Because log_3(x) is the power you'd have to raise 3 to in order to get x; and then you raise 3 to that power, so you get x.

Thanks, this part is what actually helped me understand it after plugging in values:

3^(log_3(x)) = x

3^(log_3(9)) = 9

3^(log_3(9) = [3^?=9]) = 9
3^(log_3(9) = 2) = 9
3^2 = 9
9 = 9

 

[tiny-tim]

hi daigo!

you seem to have difficulty in understanding equations

some equations are definitions, and you just have to learn them

here's a formula that may help in this case, and is easy to visualise and remember …

$\log_ax = \frac{\log x}{\log a} = \frac{ln(x)}{ln(a)} = \frac{\log_bx}{\log_ba}$ for any bases a and b

(and, as you know, a^log_ax = x by definition)

 

[genericusrnme]

If it helps, think about the log function as what it is - the inverse of exponentiation, by that I mean;
if we define f(x)=10^x, then log(x)=f^{-1}(x), and by the definition of an inverse f^{-1}(f(x)) = x

To give you a few examples take;
f(2) = 10^2 = 1000
so log(1000) = f^{-1}(1000) = 2

f(3.5) = 10^{3.5} = 3162.28
so log(3162.28) = f^{-1}(3162.28) = 3.5

When I started learning about logs understanding that it's just an inverse is what got it to 'click' for me. It's then easy to remember the things like log[a b] = log[a]log since they just come straight from the fact that 10^a 10^b = 10^(a+b)

 

[HallsofIvy]

For any positive number, a, log_a(a^x)= x and a^{log_a(x)}= x. Those essentially follow from the fact that "log_a(x)" and "a^x[/itex] are inverse functions.

 

[HallsofIvy]

If it helps, think about the log function as what it is - the inverse of exponentiation, by that I mean;
if we define f(x)=10^x, then log(x)=f^{-1}(x), and by the definition of an inverse f^{-1}(f(x)) = x

To give you a few examples take;
f(2) = 10^2 = 1000
so log(1000) = f^{-1}(1000) = 2

I'm sure this was a typo but 10^2= 100, not 1000.

f(3.5) = 10^{3.5} = 3162.28
so log(3162.28) = f^{-1}(3162.28) = 3.5

When I started learning about logs understanding that it's just an inverse is what got it to 'click' for me. It's then easy to remember the things like log[a b] = log[a]log since they just come straight from the fact that 10^a 10^b = 10^(a+b)

 

[genericusrnme]

I'm sure this was a typo but 10^2= 100, not 1000.

yes haha, I was pretty tired when I typed that