Question

log5 (x)= 4logx (5) , evaluate x.

Answer 1

What do you mean?
\[\Large \log_5 x=4 \log_x5\]?

Answer 2

yup

Answer 3

Do you know the base change formula?

Answer 4

ya

Answer 5

So can i write this like
\[\Large \log_x5 =\frac{\log_55}{\log_5x}\]

Answer 6

Do you know the value of \(\Huge \log_5 5\) ?

Answer 7

1?

Answer 8

yep,correct so now our equation reduces to :
\[\Large \log_5 x=\frac{4}{\log_5x}\]
Can you solve now?

Answer 9

\[\Large (\log_5 x)^2=4\]
Take sqrt on both sides etc
can you solve?

Answer 10

@kY_Tan

Answer 11

log5 (x)= 2? but how to convert that log5?

Answer 12

Nope,
\[\Large \log_5x=\pm 2\]

Answer 13

it can be -2 also...

Answer 14

Now you gotta take antilog on both sides so
raise both the sides to the power 5.

Answer 15

\[\Huge 5^{\log_5x}=5^{2}\]

Answer 16

every number after sqrt will be +and - together?

Answer 17

yes mostly

Answer 18

this is one solution,the other will be if you consider log(5)x=-2

Answer 19

oh now i know , so the next step is log5 (x) = 2, right?

Answer 20

@DLS

Answer 21

i didn't get you? i did all the steps

Answer 22

5^log5 (x) = 5^2
x=?

Answer 23

you know on numerator you will get x and RHS is 25

Answer 24

\[\Huge 5^{\log_5x}=x\]

Answer 25

2nd solution you will get on equating with -2

Answer 26

but why 25 is equal to x? the left hand side still have 5^log5(x) , sorry i don't understand this

Answer 27

@DLS

Answer 28

didn't you just said log base 5 (5) =1 similarly 5 to the power log base 5 (x)= x

Answer 29

sorry, honestly i still don't understand how 5^log5 x is equal to x

Answer 30

is it because 5 is an antilog then when it power a log 5 will be 1?

Answer 31

its a property

Answer 32

ok now i understood it , thanks @DLS

Answer 33

yw :)

Answer 34

Another way to do it:

\(y = \log_5(x)\) is equivalent to \(x = 5^y\)

\(y = 4\log_x(5)\) is equivalent to \(x^y = 5^4\)

So we can work with \(x = 5^y\) and \(x^y = 5^4\)

Since \(x = 5^y\) we substitute it in to the other equation to get:
\((5^y)^y = 5^4\)
\(5^{y^2} = 5^4\)
So y = 2

Which means
\(x = 5^2\)

Answer 35

@Hero can i use this method to this question? "log2 (x)+ logx (4)=3"

Answer 36

Not sure if it will work for this one.

Answer 37

alright :)

Answer 38

Is this the question?
\[\Large \log_2 x + \log _x 4 =3 \]
Please open a seperate one .

Answer 39

And please, if at all possible, write the bases of logs properly: