Question

solve for x: logx(3)^1/2 + logx 3^5 + logx 1/27 = 5/4. Can somebody help me.....

Answer 1

Is this your question?
\[\log_x {3^{\frac 12}}+\log_x {3^5}+\log_x \frac{1}{27}\ \ \ \ \ \ =\frac 5 4\]

Answer 2

yes

Answer 3

do you know the addition property of logarithms ?
\[\log_x a +\log_x b=??\]

Answer 4

yes, it's logx ab right..

Answer 5

good :)
so here we have
\[\log_x {3^{\frac 12}}+\log_x {3^5}+\log_x \frac{1}{27}\ \ \ \ \ \ =\frac 5 4\]
\[\log_x {3^{\frac 12}}+\log_x {3^5}+\log_x {3^{-3}}\ \ \ \ \ \ =\frac 5 4\]

Answer 6

Could you use the addition property to simplify the left side?

Answer 7

do I multiply the same base? logx (3^1/2)(3^5)(3^-3)

Answer 8

Base is same, so we can use the addition property
\[\log_x {3^{\frac 12}3^{5}3^{-3}}=\frac 54\]

Answer 9

so it would be logx 3^5/2

Answer 10

good, now use the fundamental property of log to find x

Answer 11

3^5/2=x^5/4 ?

Answer 12

can you create power of 5/4 on the left side

Answer 13

how do i do that... :S

Answer 14

\[\huge 3^{\frac 52}=3^{\frac {5\times 2}{2\times 2}=(3^2)^{\frac 54}}\]

Answer 15

why are we making the power the same.. i dont understand sorry im abit slow

Answer 16

\[2^4=x^4\]
x=2
we need same powers so that we could find x :)

Answer 17

Very easy question we make powers same to compare and make our work easy

Answer 18

\[\huge ({3^2})^{\frac 54}=x^{\frac 54}\]
so x= ??

Answer 19

9?

Answer 20

correct

Answer 21

yes, it's correct.
Do you understand this?

Answer 22

yes now i do, but usually I cancel the number not the power so it was kind of confusing..

Answer 23

THANK YOU :D

Answer 24

indices dont follow same rules of basic maths

Answer 25

oh it's allowed to cancel the power too?

Answer 26

if you have same bases for eg

Answer 27

\[3^2=3^2\]

Answer 28

or 3^9=y^9 ...
so y=3?

Answer 29

like that..?

Answer 30

yes correct now ur clear

Answer 31

yes thanks you :D

Answer 32

thank*

Answer 33

welcome for which exam are you studying

Answer 34

final exam for Foundation in Economics :)

Answer 35

I am in high School studying for India's most prestigious exam in India IIT

Answer 36

wow your last year in high school?

Answer 37

No second last in India we call it as FYJC just search for IIT in google

Answer 38

it's showing some kind of higher education thing.

Answer 39

No its graduation BTech

Answer 40

ohh okay sounds hard tho. All the best! :D

Answer 41

Thanks for u also all the Best wishes

Answer 42

der on fb

Answer 43

Thank you I need that :)