2 years ago

What value of x satisfies this equation? log2 (6x – 8) – log, 2^8=1 The value of x is?

Mathematics

1 answer:

Maslowich2 years ago

3 0

Answer:

$x=4$

Step-by-step explanation:

$log_2(6x-8)-log_2(8)=1\\\\log_2(\frac{6x-8}{8})=1\\ \\\frac{6x-8}{8}=2\\ \\6x-8=16\\\\6x=24\\\\x=4$

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The ratio of the sides of 2 cubes is 2 to 7. If the volume of the smaller cube is 32 u3, then the volume of the larger cube is _

satela [25.4K]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\$

$\bf \cfrac{small}{large}\qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{2}{7}=\cfrac{\sqrt[3]{32}}{\sqrt[3]{v}}\implies \cfrac{2}{7}=\sqrt[3]{\cfrac{32}{v}}\implies \left( \cfrac{2}{7} \right)^3=\cfrac{32}{v}
\\\\\\
\cfrac{2^3}{7^3}=\cfrac{32}{v}\implies v=\cfrac{7^3\cdot 32}{2^3}$

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3 years ago

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Can someone help me with this?

LekaFEV [45]

The constant variation for the relationship being shown is 4