Finding $\frac{3x+y}{6x-1} = ?$ $$2^{3x} = 18$$
$$2^y = 9$$
$$\frac{3x+y}{6x-1} = ?$$
Let me show my attempt:
$$2^y = 9, 2^y = 3^2, y = 1$$
$$2^{3x} = 18, x = 1$$
I think I've gone too wrong
 A: Hint 
$$2^{3x}=18=2\times 9=2\times 2^y=2^{y+1} \implies y=3x-1$$
Can you finish ?
A: $$2^{3x} = 18$$
$$2^y = 3^2$$
$\mapsto 2^{y+1} = 9 * 2 =18$

This means that:

$2^{y+1} = 2^{3x}$

Comparing exponents (due to same base on both sides of $=$)

$y+1=3x$
$\mapsto y = 3x-1$

We have to find:

$$\frac{3x+y}{6x-1}$$

Substituting value of $y$

$= \frac{3x + (3x-1)}{6x-1}$
$= \frac{6x-1}{6x-1}$
$=1$
$$\therefore \frac{3x+y}{6x-1} = 1$$
A: $3x=\log_218$ and $y=\log_29$.
$3x+y=\log_2(18\times9)$
$6x-1=2\log_218-1=\log_2(18\times18\div2)=\log_2(18\times9)$
$\displaystyle \frac{3x+y}{6x-1}=1$
A: By the given $$x=\frac{1}{3}\log_218$$ and $$y=\log_29.$$
Thus, $$\frac{3x+y}{6x-1}=\frac{\log_218+\log_29}{2\log_218-1}=\frac{\log_2(18\cdot9)}{\log_2\frac{18^2}{2}}=\frac{\log_2162}{\log_2162}=1.$$
A: Why not at least try to do this quite simply, using the rules for powers:
$$2^{3x+y}=2^{3x}\times 2^y=18\times 9$$
$$2^{6x-1}=\frac 12 2^{6x}=\frac 12 (2^{3x})^2=\frac 12(18)^2$$From there it is easy to conclude, and if it hadn't been, you might have got some insight into what would be making it difficult.
A: You can get the value of $x$ by doing :
$2^{3x} = 18 \implies \ln(2^{3x})=\ln(18) \implies 3x \cdot \ln(2)= \ln(18) \implies 3x = \frac{\ln(18)}{\ln(2)} \implies x=\frac{\ln(18)}{3 \cdot \ln(2)}$
And similarly, the value of $y$ by doing:
$2^y = 9 \implies \ln(2^y) = \ln(9) \implies y \cdot \ln(2) = \ln(9) \implies y= \frac{\ln(9)}{\ln(2)}$
And then simply substitute
$$\frac{3x+y}{6x-1} = \frac{3 \cdot\frac{\ln(18)}{3 \cdot \ln(2)}+\frac{\ln(9)}{\ln(2)}}{6 \cdot \frac{\ln(18)}{3 \cdot \ln(2)}-1}=\frac{\frac{\ln(18)+\ln(9)}{\ln(2)}}{\frac{2 \ln(18)-\ln(2)}{\ln(2)}} = \frac{\ln(18) + \ln(9)}{2\cdot \ln(18) - \ln(2)} = \frac{\ln(18 \cdot 9)}{\ln(\frac{18^2}{2})} = \frac{\ln(162)}{\ln(162)} = 1.$$
