Question

LOG , please help me

Answer 1

\[\huge 3^{\log27}(a ^{2}+1)^{3}\]
I know it is easy but i don't know why i am not able to simplify it

Answer 2

does it help to know that \(27=3^3\)?

Answer 3

Yeah did that

Answer 4

i guess is should first ask, what is the base of that logarithm?

Answer 5

Sorry it is 27

Answer 6

actually base ten is common log , natural log is base \(e\)

Answer 7

27 is the base

Answer 8

So we can write it as 3^3

Answer 9

\[\huge 3^{\log_{27}(a^2+1)^3}~~?\]

Answer 10

YES

Answer 11

ooh i am going to make a guess
my gues is that it is
\[\large 3^{\log_{27}(a^2+3)^2}\]

Answer 12

@SithsAndGiggles that's twice now

Answer 13

Sorry i have to do some latex practice

Answer 14

lol Satellite (a^2 + 1)^3 ...

Answer 15

quick on the draw haha

Answer 16

my typing skillz got shot off in the war

Answer 17

No.name it should be easy for you now...

\(\large{ a^{\log_{a} x} = x}\)

Answer 18

\[\huge 3^{\log_33}(a ^2+1)^{3}\]

Answer 19

\(\LARGE{ 3^{\log_{3^3} (a^2+1)^3}}\)

Remember that : \(\LARGE \log_{a^x} (y) = \cfrac{1}{x} \log_{a} y\)

Answer 20

A^2+1

Answer 21

\[\Huge a ^{2}+1\]

Answer 22

I don't i why i forgot that property lol

Answer 23

So, that base has 3 to the power of 3...

the power will come out as fraction : 1/3

That is :

\(\LARGE 3^{\cfrac{1}{3}\log_{3} (a^2+1)^3 }\)

Moreover, the power of(a^2+1) is 3... so that will also come out

\(\LARGE 3^{\cfrac{3}{3} \log_{3} (a^2+1) }\)

Answer 24

Oh yes, you got it ... it is \(a^2+1\) ..

Answer 25

I forgot that property lol

Answer 26

:) It happens. Just try to practice more questions of logarithm, and you will soon become an EXPERT! :)

Answer 27

YUP!

Answer 28

\[\huge 7^{4\log_{49}^ {a}}\]

Answer 29

For this we again write 49 as 7^7 right?

Answer 30

@mathslover

Answer 31

7^2

Answer 32

Oh yes sorry typo , 7^2
So, the whole term would become a^2

Answer 33

Yeah.