More Logarithims. Is my first method correct?

[lionely]

If log 12 = a and log 18 = b express log 2 in terms of a and b

Is this method correct?

log 12= 2log2 + log 3 = a
log 2 + 2log 3 = b

log 3= (b-log2)/2
2log2 + (b-log2)/2 = a

(4log2+b-log2)/2 = 2a/2

3log2= 2a-b
log2= (2a-b)/3


and express log 13.5 in terms of a and b, I'm not sure how to attempt this one.

One last one. If log_a x =m and log_x²a=n, find the relation between m and n.

Could I take the log of the 2nd on in base a

so log_an= x^2 or something like that?

 

[symbolipoint]

Your first method (finding log of 2 ) is correct.

 

[symbolipoint]

Second part: log 13.5 = log(27/2) ?
= log((3^3)/2)
=log(3^3)-log2
=3*log3 - log2

 

[HallsofIvy]

Yes, that is correct.

 

[SammyS]

If log 12 = a and log 18 = b express log 2 in terms of a and b

Is this method correct?

log 12= 2log2 + log 3 = a
log 2 + 2log 3 = b

log 3= (b-log2)/2
2log2 + (b-log2)/2 = a

(4log2+b-log2)/2 = 2a/2

3log2= 2a-b
log2= (2a-b)/3

and express log 13.5 in terms of a and b, I'm not sure how to attempt this one.

One last one. If log_a x =m and log_x²a=n, find the relation between m and n.

Could I take the log of the 2nd on in base a

so log_an= x^2 or something like that?

For the last one, if you do as you suggest, you get $\displaystyle \ \log_a(n)=\log_a\left( \log_{x^2}(a) \right)\,,\$ which isn't helpful.

Instead, you can use the change of base formula to change the logarithm on the left hand side to base a.

But I would be inclined to write the exponential version of $\displaystyle \ \log_{x^2}(a)=n\,,\$ then take log_a of both sides of that result.

 

[lionely]

Ohhhhhh thank you guys!

 

[lionely]

log_a x = m and log x² a = n
How how about this?

(x²)ⁿ = a

log _a x² = log _a x

Wait no

log _a (x^2)ⁿ = log _a a

so 2nlog_a x = 1

so in terms of m and n

2mn = 1

 

[HallsofIvy]

log_a x = m and log x² a = n

Do you mean $log_{x^2} a= n$?

How how about this?

(x²)ⁿ = a

log _a x² = log _a x

Wait no

log _a (x^2)ⁿ = log _a a

so 2nlog_a x = 1

If you meant $log_{x^2} a= n$, yes that is true. $log_{x^2} a= n$ is the same as $(x^2)^n= x^{2n}= a$ and, taking the logarithm, with respect to a, of that, $2n log_a(x)= 1$.

so in terms of m and n

2mn = 1

Yes, that is correct.

By the way, it is better to post new questions in new threads. Many people, seeing that a question has been answered, will not respond to that thread again.