Question

Sabendo-se que log2=0,30, log3=0,48 e 12^x=15^y, então a razão x/y é igual a

Answer 1

Propriedades utilizadas:

$\boxed{\boxed{log_{x}(a\cdot b\cdot c\cdot...\cdot z)=log_{x}(a)+log_{x}(b)+...+log_{x}(z)}}\\\\\\\boxed{\boxed{log_{x}\left(\dfrac{a}{b}\right)=log_{x}(a)-log_{x}(b)}}\\\\\\\boxed{\boxed{log(10)=log_{10}(10)=1}}$
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$12^{x}=15^{y}$

Aplicando log (base 10) nos dois lados da equação:

$log(12^{x})=log(15^{y})\\\\x\cdot log(12)=y\cdot log(15)\\\\x\cdot log(3\cdot2\cdot2)=y\cdot log(3\cdot\frac{10}{2})\\\\x\cdot\left[log(3)+log(2)+log(2)\right]=y\cdot\left[log(3)+log(10)-log(2)\right]\\\\x\cdot(0,48+0,3+0,3)=y\cdot(0,48+1-0,3)\\\\x\cdot1,08=y\cdot1,18$

Então:

$\dfrac{x}{y}=\dfrac{1,18}{1,08}\\\\\\\dfrac{x}{y}=\dfrac{118}{108}\\\\\\\boxed{\boxed{\dfrac{x}{y}=\dfrac{59}{54}}}$