Logarithms (1 Viewer)

michaeljennings · Aug 11, 2011

Could someone tell me how to do this question.

Solve for x and y
log x + log y = 6 where the log is base 3
log x - log y = 4 where the log is base 2

Hermes1 · Aug 11, 2011

michaeljennings said:
Could someone tell me how to do this question.

Solve for x and y
log x + log y = 6 where the log is base 3
log x - log y = 4 where the log is base 2

posting solution in a second.....

michaeljennings · Aug 11, 2011

thanks man!

Hermes1 · Aug 11, 2011

$\log _{3}x + \log _{3}y=6$
$\log _{3}xy = 6$
$\frac{\log_{e}xy}{\log_{e}3} = 6$ - using change of base formula
$\log_{e}xy=6\log_{e}3$
$\log_{e}xy=\log_{e}3^{6}$
$\log_{e}xy=\log_{e}729$
therefore xy = 729
similarly
$\log _{2}x-\log _{2}y = 4$
$\log _{2}\frac{x}{y} = 4$
$\frac{\log_{e}\frac{x}{y}}{\log_{e}2}=4$
$\log_{e}\frac{x}{y}=4\log_{e}2=\log_{e}16$
$\frac{x}{y}=16$

the solve from here for x and y
$x = 16y$
$16y^{2}=729$
$y^{2}=\frac{729}{16}$
therefore y = 27/4
$xy = 729$
$\frac{27x}{4}=729$
therefore:
$x = 108$
$y = \frac{27}{4}$

michaeljennings · Aug 11, 2011

ahh nice thanks a lot man

Hermes1 · Aug 11, 2011

Hermes1 said:
$\log _{3}x + \log _{3}y=6$
$\log _{3}xy = 6$
$\frac{\log_{e}xy}{\log_{e}3} = 6$ - using change of base formula
$\log_{e}xy=6\log_{e}3$
$\log_{e}xy=\log_{e}3^{6}$
$\log_{e}xy=\log_{e}729$
therefore xy = 729
similarly
$\log _{2}x-\log _{2}y = 4$
$\log _{2}\frac{x}{y} = 4$
$\frac{\log_{e}\frac{x}{y}}{\log_{e}2}=4$
$\log_{e}\frac{x}{y}=4\log_{e}2=\log_{e}16$
$\frac{x}{y}=16$

the solve from here for x and y
$x = 16y$
$16y^{2}=729$
$y^{2}=\frac{729}{16}$
therefore y = 27/4
$xy = 729$
$\frac{27x}{4}=729$
therefore:
$x = 108$
$y = \frac{27}{4}$

man i luv latex

michaeljennings · Aug 11, 2011

Hermes1 · Aug 11, 2011

michaeljennings said:
huh?

latex is the code thing that allows me to write the solutions in such a cool font.

michaeljennings · Aug 11, 2011

ohh right hahah

Hermes1 · Aug 11, 2011

Drongoski said:
$log_3 x + log_3 y = 6 \dots (1)$

$log_3 x - log_3 y = 4 \dots (2)$

drongoski why do you have base 3 for the second one when it is base 2?

Drongoski · Aug 11, 2011

Hermes1 said:
drongoski why do you have base 3 for the second one when it is base 2?

Oh really - bad eyesight! So please ignore my solution. Thanks for pointing out, Hermes1.

New Solution

$\therefore log_3 xy = 6 \implies xy = 3^6 . . . . . (1)$

$log_2 \frac {x}{y} = 4 \implies \frac {x}{y} = 2^4. . . . . (2)$

$\therefore xy \times \frac {x}{y} = x^2 = 3^6 \times 2^4 \implies x = 3^3 \times 2^2 = 108$

$xy \div \frac {x}{y} = y^2 = \frac {3^6}{2^4} \implies y = \frac {3^3}{2^2} = \frac {27}{4}$

largarithmic · Aug 11, 2011

Is this question from a past SGS trial? It looks very familiar.

HyperComplexxx · Aug 11, 2011

yea its from 2010 SGS trial

Logarithms (1 Viewer)

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