Yohan C. answered • 03/13/15

Tutor

4
(1)
Math Tutor (up to Calculus) (not Statistics and Finite)

Hey Angela,

log

_{a}b = x or a^{x}= bchange -of-base formula will be:

log

_{a}b= (log_{n}b) / (log_{n}a) n is the new base.Let's say 3

^{x}= 7 or log_{3}7 = xTake log both sides from first equation, and you will get log 3

^{x}= log 7.Bring the x (exponent or power) to the front, and you will get: x (log 3) = (log 7)

Then x = (log 7) / (log 3). Remember: If there is no base, it will be base 10. From this representation, 10 became new base.

Here is an example with integers in it:

So, let's go back to your equation log

_{a}b = xLet's rewrite as exponential equation: a

^{x}= bTake log both sides and you will get log a

^{x}= log b.Bring x (exponent or power) to the front and you will get x (log a) = (log b).

Then, your x = (log b) / (log a) (base for this log is 10)

Again, log

_{a}b = xLet a = 3, b = 81, x = 4 log

_{3}81 = 4log

_{10}81 / log_{10}3 = 4Rewrite ln function 81 with 3

^{4}and 3 with 3^{1}: log_{10}3^{4}/ log_{10}3^{1}= 4Guess what? you can bring those powers to the front and rewrite as

4 (log

_{10}3) / 1 (log_{10}3) = 4 (as log_{10}3 cancels out)I'm going to use e as a base to make ln (easy to write)

log

_{e}81 / log_{e}3 = 4 ln 81 / ln 3 = 4Rewrite ln function 81 with 3

^{4}and 3 with 3^{1}: ln 3^{4}/ ln 3^{1}= 4Guess what? you can bring those powers to the front and rewrite as

4

**(ln 3)**/ 1**(ln 3)**= 4 (as**ln 3**cancels out)I hope you understand / understood

**"Change-Base-Formula"**from this.Good luck to you.