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Difference between revisions of "2022 AMC 12A Problems/Problem 14"

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-bluelinfish
 
-bluelinfish
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===Solution 1.1===
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(Elaboration & motivation behind Sol. 1)
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Note that <math>5,20,8</math>, and <math>0.25</math> are all products and quotients of exponents of <math>2</math> and <math>5</math>, and the base of the logarithms is <math>10 = 2\times5</math>; this strongly hints at some sort of major simplification using the addition and subtraction rules of logarithms so we can convert all the different arguments of the logs into 1 common argument for easy algebra.
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Note that we can write all of the following expressions in the following ways:
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\begin{align*}
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\log5=\log\dfrac{10}2=\log10-\log2&=1-\log2\\ \log20=\log(2\cdot10)=\log2+\log10&=\log2+1\\
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\log8=\log\left(2^3\right)&=3\log2 \\
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\log0.25=\log\left(2^{-2}\right)&=-2\log2
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\end{align*}
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Thus, let <math>a=\log2</math>, and proceed as in solution 1.
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~Technodoggo
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~some elaboration by rawr3507
  
 
==Solution 2==
 
==Solution 2==
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<cmath>=2(x+y)^2 = \boxed{2}</cmath>
 
<cmath>=2(x+y)^2 = \boxed{2}</cmath>
  
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~Hithere22702
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~kempwood
  
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==Solution 3==
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==Solution 3 (Estimates)==
  
 
We can estimate the solution. Using <math>\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9</math> and <math>\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,</math> we have
 
We can estimate the solution. Using <math>\log(2) \approx 0.3, \log(20) = \log(10)+\log(2) = 1 + 0.3 \approx 1.3, \log(8) \approx 0.9</math> and <math>\log(.25) = \log(1)-\log(4)= 0 - 0.6\approx -0.6,</math> we have
  
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<cmath>0.7^3 + 1.7^3 + .9\cdot(-0.6) = \boxed{2}</cmath>
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<cmath>0.7^3 + 1.3^3 + .9\cdot(-0.6) = \boxed{2}</cmath>
 
~kxiang
 
~kxiang
  
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==Video Solution (Speedy)==
 
==Video Solution (Speedy)==
 
https://www.youtube.com/watch?v=pai2A9FXI9U
 
https://www.youtube.com/watch?v=pai2A9FXI9U
 +
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 +
==Video Solution (Simple)==
 +
https://youtu.be/7yAh4MtJ8a8?si=9vbP5erdxlCLlG82&t=2957
 +
 +
~Math-x
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}}
 
{{AMC12 box|year=2022|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:14, 5 November 2024

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 $5$, and the base of the logarithms is $10 = 2\times5$; this strongly hints at some sort of major simplification using the addition and subtraction rules of logarithms so we can convert all the different arguments of the logs into 1 common argument for easy algebra.

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 ~some elaboration by rawr3507

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

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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