logarithms again?

Question

find y in terms of x

2 log y-4 to the base 3 = 3 log x+2 to the base 3

answer is y = (100/square root x ) - 1

Answer 1

2 log[base 3](y - 4) = 3log[base 3](x + 2)

First, move all constants in front of the log to the inside of the log as an exponent. {This is due to the log property
c log[base b](a) = log[base b](a^c) }

log[base 3]( [y - 4]^2 ) = log[base 3]( [x + 2]^3 )

Now, take the antilog of both sides; this effectively eliminates each log.

(y - 4)^2 = (x + 2)^3

Taking the square root of both sides, we get

y - 4 = +/- (x + 2)^(3/2)

y = 4 +/- (x + 2)^(3/2)

Answer 2

Do you mean:
2 log₃( y - 4 ) = 3 log ₃( x + 2 )?
If yes, here's the steps:

2 log₃( y - 4 ) = 3 log ₃( x + 2 )
log₃ ( y - 4 )² = log₃( x + 2 )³
( y - 4 )² = ( x + 2 )³
y - 4 = ( x + 2 )to the power of 3/2
y = {( x + 2 )to the power of 3/2} =4

Haha... I didn't get your answer...

Answer 3

log(y-4)^2-log(x+2)^3 = 0 log[(y-4)^2/(x+2)^3=0] If the log is 0 the number is 1
(y-4)^4 =(x+2)^3 which doesn't conduct to your answer.

Lets read it another way

logy^2 -logx^3 =6
y^2/x^3 = 3^6 y^2 =3^6*x^3 and y = 3^3(x^3/2)
You must write your question with parenthesys

Answer 4

Get to hell...no one uses logarithms now except the Greeks...and no one did when i learnt them at school...what a waste of time..

Logarhthimos...greek for The Bill..bet you didnt know that..
so
2 Amstels(large @ 2Euoseach)
3 Moussakas (@7each)
3 tiramisou...( 3eu each)

35Eu including tip....y..cos its bloody good value for moneyx3
In France..try 'Addition....
Spain..La cuenta..(the count)..
Ah..Mr Edwards would be so proud of me now if he read this,,,,