# Solve simultaneously..? x = 3^y and x = 1/2 (3 + 9y)

Nov 23, 2017

This is the method I used in deducing the following simultaneously equation..

See steps below;

#### Explanation:

Solving simultaneously..

$x = {3}^{y} - - - - - - e q n 1$

$x = \frac{1}{2} \left(3 + 9 y\right) - - - - - - e q n 2$

Look at the common value in both equations..

$x$ is the common, hence we equate both together..

Having..

${3}^{y} = \frac{1}{2} \left(3 + 9 y\right)$

${3}^{y} = \frac{3 + 9 y}{2}$

Cross multiplying..

${3}^{y} / 1 = \frac{3 + 9 y}{2}$

$2 \times {3}^{y} = 3 + 9 y$

${6}^{y} = 3 + 9 y$

Log both sides..

$\log {6}^{y} = \log \left(3 + 9 y\right)$

Recall the law of logarithm $\to \log {6}^{y} = x , y \log 6 = x$

Therefore...

$y \log 6 = \log \left(3 + 9 y\right)$

Divide both sides by $\log 6$

$\frac{y \log 6}{\log 6} = \log \frac{3 + 9 y}{\log 6}$

$\frac{y \cancel{\log 6}}{\cancel{\log 6}} = \log \frac{3 + 9 y}{\log 6}$

$y = \frac{\log \left(3 + 9 y\right)}{\log} \left(6\right)$

$y = \frac{\cancel{\log} \left(3 + 9 y\right)}{\cancel{\log} \left(6\right)}$

$y = \frac{3 + 9 y}{6}$

Cross multiplying..

$\frac{y}{1} = \frac{3 + 9 y}{6}$

$6 \times y = 3 + 9 y$

$6 y = 3 + 9 y$

Collect like terms

$6 y - 9 y = 3$

$- 3 y = 3$

Divide both sides by $- 3$

$\frac{- 3 y}{- 3} = \frac{3}{-} 3$

$\frac{\cancel{- 3} y}{\cancel{- 3}} = \frac{3}{-} 3$

$y = - \frac{3}{3}$

$y = - 1$

Substitute the value of $y$ into $e q n 1$ to get $x$

$x = {3}^{y} - - - - - - e q n 1$

$x = {3}^{-} 1$

Recall in indices, ${x}^{-} 1 = \frac{1}{x}$

$\therefore x = \frac{1}{3}$

Hence the values are $\Rightarrow x = \frac{1}{3} , y = - 1$

Hope this helps!