Question

than given lg 4 = p rewrite lg 25 in function of p (using p)
lg mean logarithm base ten

i ve got lg 25 = ((2-p)/p)lg 4

so what proven

Answer 1

proven will be

((2-p)/p)lg 4 = ((2-p)/lg 4)lg 4 where lg 4 with lg 4 from denominator simplified and will remain there (2-p) so lg 25 = 2 -p

lg 25 = 2 -lg 4

lg 25 = 2lg 10 -lg 4

lg 25 = lg 10^2 -lg 4

lg 25 = lg 100 - lg 4

lg 25 = lg (100/4)

lg 25 = lg 25

hope this is right sure

Answer 2

@ganeshie8
@TheSmartOne
@mathmate
@Kainui

your opinion about this please ? ty

Answer 3

agent0smith how you see it please ?

Answer 4

If \[\log 4 = p\]then \[\large \log 25 = \frac{ \log 25 }{ \log 4 }*p\]since p=log 4.

Answer 5

yes i understand your idea but there on the right side dont use just lg 4 or p because you need rewriting the lg 25 in function of p

Answer 6

using the properties of logarithms

lg 4 = p => 10^p = 4 - yes ?

(2^p)*(5^p) = 2^2

5^p = (2^2)/(2^p)

5^p = 2^(2-p) make radical index p from both sides or push both sides on power of 1/p

5 = (2^(2-p))^(1/p)

5 = 2^((2-p)/p) squared both sides

5^2 = 2^(((2-p/p))*2)

5^2 = 4^((2-p/p) lograith lg on both sides

lg 25 = ((2-p/p)) lg 4 but we know that p=lg 4

lg 25 = ((2-lg 4)/lg 4)lg 4

lg 25 = (2-lg 4)

lg 25 = lg 10^2 -lg 4

lg 25 = lg 100 - lg 4

lg 25 = lg (100/4)

lg 25 = lg 25

i think this is right in this way

Answer 7

@ShadowLegendX
@MrNood

opinions please

ty

Answer 8

@sweetburger

Answer 9

Yeah that looks fine
\[\large \log 25 = \frac{ 2-p }{ p } *p\]or
log 25 = 2 - p

Answer 10

thank you and i think this is unbeliveblie but i solved it yestoday in my workplace dont using any from so without pen,paper or calculator and today when i ve said this to my student collage there have said me that i m crazy bc. how is this possible ? but when i ve said him this result of ((2-p)/p)lg 4 said me that is very very nice

Answer 11

your 'proof' is a trivial exercise (I mean mathematically - not effortwise1)

Answer 12

ty your opinion Nood

Answer 13

a=a is self evident

I expect your proof is 'circular - i.e. you derived 1 expression and then went 'back the same way

Answer 14

25 = 100/4

log 25 = log 100 - log 4

log 25 = 2 -p

Answer 15

ok. but you know a different way to rewrite lg 25 in function of p when you know that p=lg 4 ?

wait and thank you

Answer 16

but please ATTENTION you need starting from lg 4 = p

Answer 17

@MrNood's method is easier. I don't buy the "rules" that you need to start from a certain expression.

Answer 18

ok. thank you

Answer 19

@zepdrix your opinion please about this ? ty

Answer 20

log 4 does not equal log 25, so cannot start from log 4.
The question is "rewrite lg 25 in function of p", which means we start with log 25.
I endorse @MrNood's approach.

Answer 21

ty.

@ganeshie8 your opinion please about these ? ty.

Answer 22

@pooja195
@Callisto

Answer 23

Here's another way similar to Mr. Nood's approach

\[\log(4) = \log(2^2) = 2\log(2)\]
\[\log(25) = \log(5^2)=2\log(5)\]
Adding these we get
\[p + \log(25) = 2(\log(2) +\log(5)) = 2\log(10) = 2\]

Thus log(25) = 2 - p

Answer 24

@kl0723 np. welcome - sorry but i not can sending to you message just in this way because my submit button was blocked and not is solved this problem again and i not can sending messages to nobody - what i ve wrote there i see it true on this way -

good luck

Answer 25

@Directrix what is your opinion about this please dear my friend ?

Answer 26

@jhonyy9
what do you actually WANT from your various 'tagged' friends?

the answer is clear above.

Answer 27

^exactly. There's no need for more input.