Question

Se A log2 15x log15 2 e B 1/25 5 calcule A-B

Answer 1

Podemos aplicar a propriedade da troca de base em A:

$A=log_{_2}15\;.\;log_{_{15}}2\\\\A=\frac{log_{_{10}}15}{log_{_{10}}2}\;.\;\frac{log_{_{10}}2}{log_{_{10}}15}\\\\A=\frac{log_{_{10}}15}{log_{_{10}}15}\;.\;\frac{log_{_{10}}2}{log_{_{10}}2}\\\\A=1\;.\;1\\\\A=1$

Para B:

$B=\frac{1}{log_{_{25}}5}\\\\log_{_{25}}5=\frac{1}{B}\\\\5=25^{\frac{1}{B}}\\\\5=\left(5^2\right)^{\frac{1}{B}}\\\\5^1=5^{\;2\;.\;\frac{1}{B}}\\\\5^1=5^{\frac{2}{B}}\\\\1=\frac{2}{B}\\\\B=2$

Assim podemos calcular:

A - B = 1 - 2

A - B = -1