Logs but no wood - part 1

Algebra Level 4

If $a = \log_{12} 27$ and $\log_6{16}= x\left(\frac{y-a}{z+a}\right)$ , find the value $x+y+z$ .

The answer is 10.

1 solution

Krishna Ramesh
May 8, 2014

$\log _{ 12 }{ 27=a }$

$\Rightarrow \quad 3\log _{ 12 }{ 3 } =a$

$\Rightarrow \quad \frac { 3 }{ \log _{ 3 }{ 12 } } =a$ $\Rightarrow \quad \frac { 3 }{ 1+2\log _{ 3 }{ 2 } } =a$

$\Rightarrow \log _{ 3 }{ 2 } =\frac { 3-a }{ 2a }$ $\Rightarrow \log _{ 2 }{ 3=\frac { 2a }{ 3-a } }$

$Now,\quad \log _{ 6 }{ 16=\quad 4\log _{ 6 }{ 2 } } =\frac { 4 }{ \log _{ 2 }{ 6 } } =\frac { 4 }{ 1+\log _{ 2 }{ 3 } } =\frac { 4 }{ 1+\frac { 2a }{ 3-a } }$

$\Rightarrow \log _{ 6 }{ 16=4\left( \frac { 3-a }{ 3+a } \right) }$

$so,\quad x=4,\quad y=3\quad and\quad z=3\quad \Rightarrow \quad x+y+z=10\\$

$log_{6}16 = 4log_{6}2 = 4 \frac{log_{12}2}{log_{12}6} = 4 \frac{log_{12}2 + a - a}{log_{12}6 - a + a}$

$\Rightarrow log_{6}16 = 4 \frac{log_{12}(2 \times 27) - a}{log_{12} \frac{6}{27} + a}$ $so,\ x = 4, \\\\ y = log_{12}(2 \times 27) \ and \\\\ z = log_{12}\frac{6}{27}$

$\Rightarrow x + y + z = 5$

What is the error in this solution?

Tushar Marda - 7 years ago

I solved the problem this way too.

Sissi Jian - 5 years, 11 months ago

excellent!

Mayank Holmes - 7 years ago

Logs but no wood - part 1

The answer is 10.

1 solution

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