Question

Log 4 X - log 8 X = 1

Log de x na base 4 menos log de x na base 8 igual a 1

O conjunto solução é 64

Answer 1

$log_4(x) - log_8(x) = 1 \\\\ log_{2^2}(x) - log_{2^3}(x) = 1 \\\\ {1 \over 2}log_2(x) - {1 \over 3}log_2(x) = 1 \\\\ {3log_2(x) - 2log_2(x) \over 6} = 1 \\\\ {log_2(x) \over 6 } = 1 \\\\ log_2(x) = 6 \\\\ x = 2^6 \\\\ \boxed{ x = 64}$

Answer 2

$\large\boxed{\begin{array}{l}\underline{\sf Consequ\hat encia\,da\,mudanc_{\!\!,}a\,de\,base}\\\rm Se\,a\,e\,b\,s\tilde ao\,reais\,positivos\,com\,a\,diferente\,de\,1\\\rm e\,\beta\,\acute e\,um\,real\,n\tilde ao\,nulo\,ent\tilde ao\,tem-se:\\\boxed{\boxed{\rm \ell og_{a^{\beta}}b=\dfrac{1}{\beta}\ell og_ab}}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm prova\!:}\\\bf 1^o~caso:~se~b=1~temos:\\\\\begin{cases}\rm \ell og_a1=0\\\rm\ell og_{a^{\beta}}1=0 \end{cases}\longrightarrow \rm \ell og _{a^{\beta}}1=\dfrac{1}{\beta}\cdot\ell og_a1\\\bf 2^o~caso: Se~b\ne1,temos:\\\rm \ell og_{a^{\beta}}b=\dfrac{1}{\ell og_ba^{\beta}}=\dfrac{1}{\beta\ell og_ba}=\dfrac{1}{\beta}\cdot\dfrac{1}{\ell og_ba}\\\\\rm \ell og_{a^{\beta}}b=\dfrac{1}{\beta}\cdot \ell og_ab~\blacksquare\end{array}}$

$\large\boxed{\begin{array}{l}\rm \ell og_4x-\ell og_8x=1\\\sf usando\,a\,propriedade\,j\acute a\,demonstrada\,temos:\\\rm \ell og_{2^2}x-\ell og_{2^3}x=1\\\\\rm\dfrac{1}{2}\ell og_2x-\dfrac{1}{3}\ell og_2x=1\bullet(6)\\\\\rm 3\ell og_2x-2\ell og_2x=6\\\rm \ell og_2x=6\\\rm x=2^6\\\rm x=64\\\rm S=\{64\}\end{array}}$