Question

1) aplicando a definição, calcule o valor dos logaritmos :​

Answer 1

$\large\boxed{\begin{array}{l}\underline{\rm D~\!\!efinic_{\!\!,}\tilde ao\,de\,logaritmo}\\\sf \ell og_ba=x\iff b^x=a\\\underline{\rm Condic_{\!\!,}\tilde ao\,de\,exist\hat encia\,de\,um\,logaritmo}\\\sf \ell og_ba\exists\iff\begin{cases}\sf a>0\\\sf b>0\\\sf b\ne1\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\rm a)~\sf\ell og_{\sqrt{8}}4=x\\\sf (\sqrt{8})^x=4\\\sf 2^{\frac{3x}{2}}=2^2\\\sf \dfrac{3x}{2}=2\\\\\sf 3x=4\\\\\sf x=\dfrac{4}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\rm b)~\sf \ell og_{25}0,2=x\\\sf 25^x=0,2\\\sf 25^x=\dfrac{1}{5}\\\sf 5^{2x}=5^{-1}\\\sf 2x=-1\\\sf x=-\dfrac{1}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\rm c)~\sf \ell og_2\sqrt[\sf3]{\sf64}=x\\\sf 2^x=\sqrt[\sf3]{\sf64}\\\sf 2^x=64^{\frac{1}{3}}\\\sf 2^x=[(2^2)^3]^{\frac{1}{3}}\\\sf 2^x=(2^2)^{3\cdot\frac{1}{3}}\\\sf 2^x=2^2\\\sf x=2\end{array}}$

$\large\boxed{\begin{array}{l}\rm d)~\sf \ell og_{16}32=x\\\sf 16^x=32\\\sf (2^4)^x=2^5\\\sf 2^{4x}=2^5\\\sf 4x=5\\\sf x=\dfrac{5}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm e)~\sf \ell og_50,000064=x\\\sf 5^x=0,000064\\\sf 5^x=\dfrac{64\div64}{1000000\div64}\\\\\sf 5^x=15625\\\sf 5^x=5^6\\\sf x=6\end{array}}$

$\large\boxed{\begin{array}{l}\rm f)~\sf\ell og_{49}\sqrt[\sf3]{\sf7}=x\\\sf 49^x=\sqrt[\sf3]{\sf7}\\\sf (7^2)^x=7^{\frac{1}{3}}\\\sf 7^{2x}=7^{\frac{1}{3}}\\\sf 2x=\dfrac{1}{3}\\\\\sf 6x=1\\\sf x=\dfrac{1}{6}\end{array}}$

$\large\boxed{\begin{array}{l}\rm g)~\sf \ell og_381=x\\\sf 3^x=81\\\sf 3^x=3^4\\\sf x=4\end{array}}$

$\large\boxed{\begin{array}{l}\rm h)~\sf \ell og_2\sqrt[\sf8]{\sf64}=x\\\sf 2^x=\sqrt[\sf8]{\sf64}\\\sf 2^x=2^{\frac{6}{8}}\\\sf x=\dfrac{6\div2}{8\div2}\\\\\sf x=\dfrac{3}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm i)~\sf \ell og_42\sqrt{2}=x\\\sf 4^x=2\sqrt{2}\\\sf (2^2)^x=2\cdot 2^{\frac{1}{2}}\\\sf 2^{2x}=2^{\frac{3}{2}}\\\sf 2x=\dfrac{3}{2}\\\\\sf 4x=3\\\sf x=\dfrac{3}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\rm j)~\sf\ell og_20,25=x\\\sf 2^x=0,25\\\sf 2^x=\dfrac{1}{4}\\\\\sf 2^x=2^{-2}\\\\\sf x=-2\end{array}}$

$\large\boxed{\begin{array}{l}\rm l)~\sf\ell og_{\sqrt[\sf5]{\sf2}}128=x\\\sf (\sqrt[\sf5]{\sf2})^x=128\\\sf 2^{\frac{x}{5}}=2^7\\\sf \dfrac{x}{5}=7\\\\\sf x=7\cdot5\\\sf x=35\end{array}}$

$\large\boxed{\begin{array}{l}\rm m)~\sf \ell og_{625}\sqrt{5}=x\\\sf 625^x=\sqrt{5}\\\sf (5^4)^x=5^{\frac{1}{2}}\\\sf 5^{4x}=5^{\frac{1}{2}}\\\sf 4x=\dfrac{1}{2}\\\\\sf 8x=1\\\\\sf x=\dfrac{1}{8}\end{array}}$