One of the solutions to the above equation would be where $a, b$ and $c$ are equal to 0 , but there is a condition that says neither of those variables are equal to each other. I have a feeling the question is wrong . The answer given at the back of the textbook is $2(a+b)c = ab$. Can somebody explain me the process
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HINT:
Let $5^a=3^b=225^c=k\implies k\ne0,1$
$\implies5=k^{1/a},3=k^{1/b}$
$k^{1/c}=225=3^2\cdot5^2=(k^{1/a})^2\cdot(k^{1/b})^2=k^{2/a+2/b}$
lab bhattacharjee
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Let $t=\log_35$. Then $b=at$.
$5^a=3^b=225^c=5^{2c}\cdot 3^{2c}$, therefore $$5^{a-2c}=3^{2c}$$ and $$3^{b-2c}=5^{2c}$$
That is $$(a-2c)t=2c$$ $$b-2c=2ct$$ These two equations, along with $b=at$ form an homogeneous linear system:
$$\begin{pmatrix}t&0&-2-2t\\0&1&-2-2t\\t&-1&0\end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}$$
Since the system matrix is not singular, it has infinitely many solutions.
The relationsship between $a$, $b$ and $c$ is $$b=a\log_35=c\log_3225$$
ajotatxe
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How this is true $t = \log_3 5$, then $b = at$? – YOUSEFY Oct 26 '16 at 02:55