Question

Plz help!!!!!!!!!
Pre-calulus

Answer 1

The value is: 1

Explanation:

Use the Change of base formula. This is:

`log_a(x)=(log_b(x))/(log_b(a))`

Using base 10:

`log_3(5)*log_(25)(9)=(log(5))/(log(3))*(log(9))/(log(25))`

We know that:

`9=3^2\\25=5^2`

And according to the logarithms properties:

`log(x)^n=nlog(x)`

Then, we can simplify the expression:

`=(log(5))/(log(3))*(log(3)^2)/(log(5)^2)=(log(5))/(log(3))*(2log(3))/(2log(5))=(log(5))/(log(3))*(log(3))/(log(5))=(log(5)*log(3))/(log(3)*log(5))=1`

Answer 2

1

Explanation:

We are to find the value of the following:

`log _ 3 5 * log _ { 2 5 } 9`

`\log _ { 3 } 5 * \log _ { 2 5 } 9 =(1)/(\log_(5)3)*\log_(25)9=(1)/(\log_(5)3)*\log_(5^2)9=(1)/(\log_(5)3)*(1)/(2)\log_(5)9`

`\frac { 1 } { \log _ { 5 } 3 } * \frac { 1 } { 2 } \log_ { 5 } 9 = (1)/(\log_(5)3)\cdot(1)/(2)\log_(5)3^2=(1)/(\log_(5)3)\cdot(2)/(2)\log_(5)3=(1)/(\log_(5)3)\cdot\log_(5)3=(\log_(5)3)/(\log_(5)3)=` 1