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Algebra Level 3

$\begin{aligned} a & = & \frac { \log_{27} 8}{\log_ 3 2} \\ b & = & \left ( \frac 1 {2^{\log_2 5}} \right ) \left ( \frac 1 {5^{\log_5 0.1}}\right ) \\ c & = & 2^{\sqrt{\log_2 5}} - 5^{\sqrt{\log_5 2}} \\ \end{aligned}$

What is the value of $a+b+c$ ?

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The answer is 3.

1 solution

Brian Charlesworth
Mar 31, 2015

Using the exponent and change of base rules, we have that

$a = \dfrac{3*\log_{27}(2)}{\log_{3}(2)} = \dfrac{3*\frac{\log_{3}(2)}{\log_{3}(27)}}{\log_{3}(2)} = 1$ and $b = \left(\dfrac{1}{5}\right)\left(\dfrac{1}{0.1}\right) = 2.$

To determine $c,$ note that $\sqrt{\log_{2}(5)} = \dfrac{\log_{2}(5)}{\sqrt{\log_{2}(5)}} = \log_{2}(5)\sqrt{\log_{5}(2)}.$ Thus

$c = \large (2^{\log_{2}(5)})^{\sqrt{\log_{5}(2)}} - 5^{\sqrt{\log_{5}(2)}} = 5^{\sqrt{\log_{5}(2)}} - 5^{\sqrt{\log_{5}(2)}} = 0.$

Therefore $a + b + c = 1 + 2 + 0 = \boxed{3}.$

What about this method

$a= \frac{log_{27}8}{log_{3}2} = \frac{log_{3^3} 2^3}{log_{3}2} =\frac{log_{3}(2^3)^\frac{1}{3}}{log_{3}2} = \frac{log_{3}2}{log_{3}2} = 1$

$b= (\frac{1}{2^{log_{2}5}})(\frac{1}{5^ {log_{5}0.1}}) = (\frac{1}{5})(\frac{1}{0.1}) = (\frac{1}{5})(10) = 2$

To find $c,$ see that $\sqrt{\log_{2}(5)} = \dfrac{\log_{2}(5)}{\sqrt{\log_{2}(5)}} = \log_{2}(5)\sqrt{\log_{5}(2)}.$

Thus

$c = (2^{\log_{2}(5)})^{\sqrt{\log_{5}(2)}} - 5^{\sqrt{\log_{5}(2)}} = 5^{\sqrt{\log_{5}(2)}} - 5^{\sqrt{\log_{5}(2)}} = 0.$

Therefore $a + b + c = 1 + 2 + 0 = \boxed{3}.$

Abhisek Mohanty - 6 years, 2 months ago

I failed to evaluate c

Vighnesh Raut - 6 years, 2 months ago

same here . I failed to calculate c

Sai Ram - 5 years, 2 months ago

Me too, failed to calculate c

Shivansh Jaiswal - 4 years, 10 months ago

Failed to calculate C

I Gede Arya Raditya Parameswara - 4 years, 4 months ago

Sometimes logs logs our mind too!

Want some more than click here

The answer is 3.

1 solution

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