Após a realização dos cálculos ✍️, podemos concluir mediante ao conhecimento de logaritmos que:

a) [tex]\sf \log_{25}0,008=-\dfrac{3}{2}[/tex] ✅

b) [tex]\sf \log_{\sqrt[3]{\sf5}}\sqrt[\sf4]{\sf5}=\dfrac{3}{4}[/tex] ✅

c) [tex]\sf \log_{\sqrt[\sf3]{\sf7}}49=6[/tex] ✅

d) [tex]\sf \log_{125}\dfrac{1}{5}=-\dfrac{1}{3}[/tex] ✅

Logaritmo

Chama-se logaritmo de  um número real b positivo, na base a positiva e diferente de 1, o expoente x a qual se deve elevar a para se obter b.

matematicamente:

[tex]\sf \log_ab=x\iff a^x=b,\begin{cases}\sf b > 0\\\sf a > 0\\\sf a\ne1\end{cases}[/tex]

✍️Vamos a resolução do exercício

Nos itens de a até d iremos utilizar a definição de logaritmo para resolver.

a)

[tex]\Large\boxed{\begin{array}{l}\sf 0,008=\dfrac{8}{1000}=\dfrac{2^3}{10^3}=\bigg(\dfrac{2\div2}{10\div2}\bigg)^3=\bigg(\dfrac{1}{5}\bigg)^3\\\\\sf \log_{25}0,008=x\\\sf 25^x=0,008\\\sf (5^2)^x=\bigg(\dfrac{1}{5}\bigg)^3\\\\\sf 5^{2x}=5^{-3}\\\sf 2x=-3\\\\\sf x=-\dfrac{3}{2}\end{array}}[/tex]

b)

[tex]\Large\boxed{\begin{array}{l}\sf \log_{\sqrt[\sf3]{\sf5}}\sqrt[\sf4]{\sf5}=x\\\\\sf(\sqrt[\sf3]{\sf5})^x=\sqrt[\sf4]{\sf5}\\\\\sf 5^{\frac{x}{3}}=5^{\frac{1}{4}}\\\sf\dfrac{x}{3}=\dfrac{1}{4}\\\\\sf 4x=3\\\\\sf x=\dfrac{3}{4}\end{array}}[/tex]

c)

[tex]\Large\boxed{\begin{array}{l}\sf\log_{\sqrt[\sf3]{\sf7}}49=x\\\sf(\sqrt[\sf3]{\sf7})^x=49\\\sf(7^{\frac{1}{3}})^x=7^2\\\\\sf 7^{\frac{x}{3}}=7^2\\\sf\dfrac{x}{3}=2\\\\\sf x=3\cdot2\\\sf x=6\end{array}}[/tex]

d)

[tex]\Large\boxed{\begin{array}{l}\sf\log_{125}\dfrac{1}{5}=x\\\\\sf 125^x=\dfrac{1}{5}\\\\\sf (5^3)^x=5^{-1}\\\\\sf 5^{3x}=5^{-1}\\\sf 3x=-1\\\\\sf x=-\dfrac{1}{3}\end{array}}[/tex]

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