Hello there. To solve this question, we'll have to apply some logarithm properties to rewrite the number and determine the values of A and B.
Given
[tex]\frac{2}{\log (20)-1}=\log _B(A)[/tex]
We can rewrite the expression in the denominator using the following rule:
[tex]\log _C(a)-\log _C(b)=\log _C\left(\frac{a}{b}\right)[/tex]
Remember that we write log for a logarithm with base 10 and ln for a logarithm with base e, thus we have:
[tex]\log (20)-1=\log (20)-\log (10)=\log \mleft(\frac{20}{10}\mright)=\log (2)[/tex]
And rewrite the expression in the numerator using the property:
[tex]\log _C(a^b)=b\cdot\log _C(a)[/tex]
Therefore we have:
[tex]2=2\cdot1=2\cdot\log (10)=\log (10^2)=\log (100)[/tex]
Finally, we use the following property to find A and B:
[tex]\frac{\log_C(a)}{\log_C(b)}=\log _b(a)[/tex]
We get that:
[tex]\frac{\log(100)}{\log(2)}=\log _2(100)[/tex]
And this is equal to:
[tex]\log _2(100)=\log _B(A)[/tex]
If only A = 100 and B = 2.
