Question

solve for x by using the exponential form: log3 (7-x)=3

Answer 1

The first (and best) rule of logs is the definition

\[log_b (a) = k \iff b^k = a\]

Answer 2

In your case, what is a, b, and k?

Answer 3

b=3

Answer 4

Correct!

Answer 5

a=3

Answer 6

not quite.

Answer 7

I'll give you a hint. k = 3.

Answer 8

oh the 7-x

Answer 9

Right.

Answer 10

So what the definition means is if you have something that looks like this:
\[log_b (a) = k\]
It means the same thing as this:
\[b^k = a\]

And vice versa.

Answer 11

So you have something that looks like this:
\[log_3(7-x) = 3\]

How would you rewrite it using the definition?

Answer 12

3^3= 7-x

Answer 13

Correct!

Answer 14

And solving from there is a breeze.

Answer 15

yepp thanks!

Answer 16

now a question in my textbook says:" use properities of logs or a definition to simpligy the expression: If f(x)= 10^x, find (log2). can you translate that for me?

Answer 17

Huh? I'm not sure I understand what it's asking for.

Answer 18

the log base 2 of 10^x ?

Answer 19

i guess that is what it i asking?

Answer 20

If f(x)= 10^x, find f(log2).

Answer 21

Oh. that just means plug in log(2) for x. and since the base isn't specified it's log base 10.

Answer 22

oh so it would be a decimal #?

Answer 23

No

Answer 24

\[f(log(2)) = 10^{log(2)}\]

Since \[log_{10} 2 = k \iff 10^k = 2\]
Then
\[10^{log(2)} = 10^k = 2\]

Answer 25

Does that make sense?

Answer 26

The take away message being if you raise the base to the power of a log you will get the input.
\[b^{log_b(x)} = x\]

Answer 27

Same for taking the log of a power..
\[log_b(b^x) = x\]

Answer 28

But if you can't remember all those rules ( I rarely do ) you can always find them from the rule I gave at the start. The definition of the log.

Answer 29

x^2= 10?

Answer 30

No no.

Lets say that \[log_{10}(2) = k\]
What would that mean by the law of logs?

Answer 31

Go back to the definition I gave at the beginning. What is a,b,and k?

Answer 32

Actually in this case k is k, so no need for that ;p