Logarithms...... Ahhhhhh

Max.....I assume you want this in terms of x and y instead of a and b

x = log₂ 3        y = log₃ 7    log₄₂ 56         use the change of base rule to write :

x = log 3 / log 2  →   log 2  =  log 3/ x

y =  log 7 / log 3   →  log 7  = y log 3      

log₄₂ 56   =    log 56 / log 42        and we can write this as :

log ( 2^3 * 7 )  / log ( 2 * 7 * 3)   =

[log2^3 + log 7] / [ log 2 + log 7 +  log 3]  =

[3 log 2 + y log 3 ]  / [  log3 / x + y log 3 + log 3]  

[ 3 * (log 3)/x + y log 3 ]  / [ log 3/x + ylog 3 + log 3]  =   [multiply top/bottom by x]

[ 3 log 3  + xylog3 ] / [ log 3 + xylog 3 + x log 3]  =      factor out log 3

 [ (log 3) ( 3 + xy ) ]  /  [ (log 3) [ 1 + xy + x ]   =       

[ 3 + xy ]  / [ 1 + xy + x ]

1) 

Solve for b:
x^2 log(a)+log(b) (x+1) = 0

Subtract x^2 log(a) from both sides:
log(b) (x+1) = -(x^2 log(a))

Divide both sides by x+1:
log(b) = -(x^2 log(a))/(x+1)

Cancel logarithms by taking exp of both sides:
Answer: |b = a^(-x^2/(x+1))

First,nice questions,MaxWong.

For (1) I want to add some step from the equation given by the guest\(log(b)=-({x}^{2}*log(a))/(x+1)=-({x}^{2}/(x+1)*log(a))=log({a}^{-\frac{{x}^{2}}{(x+1)}})\)

Now,cancel logarithms by taking exp with both sides:

\(b={a}^{-\frac{{x}^{2}}{(x+1)}}\)

(2)I assume you mean \(a=log{2}^{3} , b=log{3}^{7}\)instead of x and y

\(log3=a*log2,log2=\frac{log3}{a}, log7=b*log3\)

\(log{42}^{56}=log{42}^{42*4/3}=1+log{42}^{4/3}=1+log(4/3)/log42=1+\frac{log(4/3)}{(log2+log3+log7)}\)

Therefore,\(log{42}^{56}=1+\frac{log(4/3)}{(log3)/a+a*log2+b*log3}\)

(3), Given that a>1 and b>1 and \({a}^{x}*{b}^{y}={a}^{y}*{b}^{x}=1\), prove that x + y = 0.

\({a}^{x}*{b}^{y}=1\Rightarrow log({a}^{x}*{b}^{y})=log1=0\)

\(x*log(a)+y*log(b)=0 \)(1)

Same for the second equation,we have

\(y*log(a)+x*log(b)=0\)(2)

combine (1) and (2),we have

\((x+y)(log(a)+log(b))=0\)(3)

Since a>1 and b>1,so log(a)+log(b)>0

So,the equation (3) is true if and only if x+y equal to 0

Hence,x+y=0 if a>1 and b>1