A 10 seconds question!

Algebra Level 2

$\large { \color{#D61F06}a^{\color{#EC7300}\log \color{#00ff33}b - \color{#BA33D6}\log \color{#3D99F6}c} \color{#624F41}\times \color{#00ff33}b^{\color{#ff0099}\log \color{#3D99F6}c - \color{#993300}\log \color{#D61F06}a} \color{#624F41}\times \color{#3D99F6}c^{\color{#000066}\log \color{#D61F06}a - \color{#ff0099}\log \color{#00ff33}b} }$

Find the value of the expression above for positive real numbers $a,b,c$ .

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0 1

\infty

\log abc

abc

5 solutions

Prasad Khole
Apr 15, 2015

I would go by standard algebraic method i.e. Y=a^(logb/c) × b^(logc/a) × c^(loga/b) Log(y)= loga×(logb-logc)+logb ×(logc-loga)+logc× (loga-logb) Logy=0 Y=1

Josh Banister
Apr 15, 2015

For this question. We will assume that we are taking logs to the base $k$ $a^{\log b - \log c} \times b^{\log c - \log a} \times c^{\log a - \log b} \\ =\frac{a^{\log b} \times b^{\log c} \times c^{\log a}}{a^{\log c} \times b^{\log a} \times c^{\log b}} \\ = \frac{k^{\log b \times \log a} \times k^{\log c \times \log b} \times k^{\log a \times \log c}}{k^{\log a \times \log b} \times k^{\log b \times \log c} \times k^{\log c \times \log a}} \\ = \boxed{1}$

I don't get the last step

Francisco Rodríguez - 6 years, 1 month ago

Note that $a=k^{\log_k a}, b=k^{\log_k b}, c=k^{\log _k c}$ Writing $\log$ instead of $\log_k$ , you go from the second to the third step (in the solution above). The last step is just cancellations.

Osama Kawish - 6 years, 1 month ago

there was no need of the k part ...

Vyom Jain - 6 years, 1 month ago

Taking the base as $k$ generalizes the solution for all positive real bases and thus, this is a better solution than simply showing it for base $10$ or base $e$ .

Prasun Biswas - 6 years, 1 month ago

Assuming $k \in \mathbb{R} - [0,1]$ .

Guilherme Naziozeno - 5 years, 11 months ago

Olamideji Akorede
Jun 7, 2015

Gamal Sultan
Apr 19, 2015

Let the given expression be denoted by M

Then

log M = (log b - log c) X log a + (log c - log a) X log b + (log a - log b) X log c = 0

M = 1

Another solution

Let log a , log b , log c be denoted by x , y , z respectively

Then

a = 10^x , b = 10^y , c = 10^z

So , the given expression = 10^x(y - z) X 10^y(z - x) X 10^z(x - y) = 10^0 = 1

Gamal Sultan - 6 years, 1 month ago

Steven Mai
Apr 17, 2015

my friend timed me and i got about 8 seconds! What i did was seperated each power (like a to the logb-logc goes to a to logb divided logc) and then i noticed that every log has its negative in the expression(logb is in a to the logb and also c to the negative logb) so the power is zero and the answer is 1

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