2022 AMC 12A Problems/Problem 14

Problem

What is the value of $(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)$ where $\log$ denotes the base-ten logarithm?

$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$

Solution 1

Let $\text{log } 2 = x$ . The expression then becomes $(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.$

-bluelinfish

Solution 1.1

(Elaboration & motivation behind Sol. 1)

Note that $5,20,8$ , and $0.25$ are all products and quotients of exponents of $2$ and $5$ ; this strongly hints at some sort of major simplification.

Note that we can write all of the following expressions in the following ways:

\begin{align*} \log5=\log\dfrac{10}2=\log10-\log2&=1-\log2\\ \log20=\log(2\cdot10)=\log2+\log10&=\log2+1\\ \log8=\log\left(2^3\right)&=3\log2 \\ \log0.25=\log\left(2^{-2}\right)&=-2\log2 \end{align*}

Thus, let $a=\log2$ , and proceed as in solution 1.

~Technodoggo

Solution 2

Using sum of cubes $(\log 5)^{3}+(\log 20)^{3}$ $= (\log 5 + \log 20)((\log 5)^{2}-(\log 5)(\log 20) + (\log 20)^{2})$ $= 2((\log 5)^{2}-(\log 5)(2\log 2 + \log 5) + (2\log 2 + \log 5)^{2})$ Let x = $\log 5$ and y = $\log 2$ , so $x+y=1$

The entire expression becomes $2(x^2-x(2y+x)+(2y+x)^2)-6y^2$ $=2(x^2+2xy+4y^2-3y^2)$ $=2(x+y)^2 = \boxed{2}$

~kempwood

Solution 3 (Estimates)

We can estimate the solution. Using $\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9$ and $\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,$ we have

$0.7^3 + 1.3^3 + .9\cdot(-0.6) = \boxed{2}$ ~kxiang

Solution 4(log bash)

Using log properties, we combine the terms to make our expression equal to $\log {( (5^{\log^2{5}}) \cdot (20^{\log^2{20}}) \cdot 8 ^ {\log{\frac{1}{4}}} ) }$ . By exponent properties, we separate the part with base $20$ to become $20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}}$ . Then, we substitute this into the original expression to get $\log {( (5^{\log^2{5}}) \cdot 20^{\log^2{5}} \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) } = \log {( (100^{\log^2{5}}) \cdot 20^{\log^2{20}-\log^2{5}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }$ . Because $100^{\log^2{5}} = 25^{\log{5}}$ , and $\log^2{20}-\log^2{5} = (\log{20}+\log{5})(\log{20}-\log{5}) = \log{100}\cdot\log{4} = 2\log{4}$ , this expression is equal to $\log {( 25 ^ {\log{5}} \cdot 400^{\log{4}} \cdot 8 ^ {\log{\frac{1}{4}}} ) }$ . We perform the step with the base combining on $25$ and $400$ to get $25 ^ {\log{5}} \cdot 400^{\log{4}} = 25 ^ {\log{5}-\log{4}} \cdot 10000^{\log{4}} = 25^{\log{\frac{5}{4}}}\cdot 256$ . Putting this back into the whole equation gives $\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 8^{\log{\frac{1}{4}}})}$ . One last base merge remains - but $25\cdot 8$ isn't a power of 10. We can rectify this by converting $8^{\log{\frac{1}{4}}}$ to $(4^\frac{3}{2})^{\log{\frac{1}{4}}} = 4^{\log{ \frac{1}{8} }}$ . Finally, we complete this arduous process by performing the base merge on $\log{( 25^{\log{\frac{5}{4}}}\cdot 256 \cdot 4^{\log{\frac{1}{8}}})}$ . We get $25^{\log{\frac{5}{4}}} \cdot 4^{\log{\frac{1}{8}}} = 25^{\log{\frac{5}{4}}-\log{\frac{1}{8}}} \cdot 100^{\log{\frac{1}{8}}} = 25^{\log{10}} \cdot \frac{1}{64} = \frac{25}{64}$ . Putting this back into that original equation one last time, we get $\log(256 \cdot \frac{25}{64}) = \log{100} = \boxed{2}$ . ~aop2014

Video Solution (Speedy)

https://www.youtube.com/watch?v=pai2A9FXI9U

~Education, the Study of Everything

Video Solution (Simple)

https://youtu.be/7yAh4MtJ8a8?si=9vbP5erdxlCLlG82&t=2957

~Math-x

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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