Question

Prove? `log_x(y)xxlog_y(x)=1`

Answer 1

See explanation.

Explanation:

The change of base formula says `log_a(b) = log(b)/log(a)`.

Applying that here we have:

`log_x(y)*log_y(b) = log(y)/log(x)\*log(x)/log(y)`

`=log(y)/log(y)\*log(x)/log(x) =1`

Answer 2

By definition of the logarithm we have:

`log_a b = c iff a^c=b`

If we apply the definition to `log_xy`; then:

`log_xy = Aiff x^A= y`

And if we apply the definition to `log_yx`; then

`log_yx =B iff y^B = x`

And if we replace `x` from the second expression into `x` from the first expression; then we gte:

`(y^B)^A= y iff y^(AB)= y iff AB =1`

And the result follows, as:

`AB =1 iff log_xy \ log_yx =B = 1` QED

Answer 3

See below

Explanation:

There are a couple of ways to do this:

• Use the log base switch rule: `log_a(b)=1/log_b(a)`

`log_x(y)xxlog_y(x)=1`

`1/log_y(x)xxlog_y(x)=log_y(x)/log_y(x)=1`

• Use the log base change rule: `log_a(b)=log_x(b)/log_x(a)`

`log_x(y)xxlog_y(x)=1`

`logy/logx xx logx/logy=(logxlogy)/(logxlogy)=1`