Changing the bases of logs question

[lionely]

Homework Statement

If x= log2aa
y= log3a2a
z= log4a3a

prove that xyz +1 = 2yz

Homework Equations

log_a C = log₁₀C/log₁₀a

The Attempt at a Solution

log= log₁₀

x= log_2aa = log a/ log2a


y= log_3a2a = log 2a/log 3a


z= log_4a3a = log 3a/log 4a


xyz + 1 = (log a/log2a)x (log 2a/log 3a) x (log 3a/log4a) + log 10


= log a/ log 4a + log 10

Is this correct so far?

 

[Curious3141]

Homework Statement

If x= log2aa
y= log3a2a
z= log4a3a

prove that xyz +1 = 2yz

Homework Equations

log_a C = log₁₀C/log₁₀a

The Attempt at a Solution

log= log₁₀

x= log_2aa = log a/ log2a


y= log_3a2a = log 2a/log 3a


z= log_4a3a = log 3a/log 4a


xyz + 1 = (log a/log2a)x (log 2a/log 3a) x (log 3a/log4a) + log 10


= log a/ log 4a + log 10

Is this correct so far?

Everything you've written there is correct. But instead of writing log 10 in the last step, I would suggest you write the "1" as \frac{\log 4a}{\log 4a} and combine the fractions.

 

[lionely]

So it's
(loga/log4a) + (log4a/log4a) = (loga/log4a) x (log4a/log4a)

= loga/log4a?

 

[Curious3141]

So it's
(loga/log4a) + (log4a/log4a) = (loga/log4a) x (log4a/log4a)

= loga/log4a?

No. How'd you get from the '+' to the 'x'?

Are you getting confused by the log rule that says: \log m + \log n = \log mn? Because that's something quite different.

Just do the addition like normal fraction expressions. The numerator of the combined expression should then be simplified using that log rule I mentioned.

 

[lionely]

so it's log4a^2/log4a?

 

[Curious3141]

so it's log4a^2/log4a?

Yes. Now observe that 4a^2 = (2a)^2.

Use \log m^n = n\log m here.

It's already quite close to the form you require. You just need one more trivial trick to introduce \log 3a into the expression. Remember that \frac{x}{y} = (\frac{x}{z})(\frac{z}{y}), where z can be anything because it just cancels out. A little bit more manipulation and you'll get the form you require.

 

[lionely]

How about (2log2a/log3a) x (log3a/log4a)?

 

[lionely]

Oh yes it is that because yz = (log3a/log4a) x (log2a/log3a)and xyz + 1 =( 2log2a/log3a) x (log3a/log4)

and that is equal to 2yz!

 

[Curious3141]

Oh yes it is that because yz = (log3a/log4a) x (log2a/log3a)


and xyz + 1 =( 2log2a/log3a) x (log3a/log4)

and that is equal to 2yz!

Yup, you got it.

 

[lionely]

Thank you!

 

[lionely]

=/ My teacher said what I did was wrong... he said it must be done the way he did it... he changed base of x to base of y then he got xy I think. Then changed the base of xy to base of z.

I don't remember exactly what he did i didn't get a chance to transcribe it... but it was something along the lines of what I said above. But I personally believe the question can be worked more than one way..

 

[haruspex]

=/ My teacher said what I did was wrong...

You have my sympathies.

 

[lionely]

What does that mean? I have a bad teacher? =/

 

[Curious3141]

What does that mean? I have a bad teacher? =/

Everything you did is mathematically valid. The only small caveat is that 'a' should be taken to be positive, but this is generally a given in this sort of question. I don't know what your teacher is on about.