$$2^{3x} = 18$$ $$2^y = 9$$ $$\frac{3x+y}{6x-1} = ?$$
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
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4How does $2^y=3^2$ make $y=1$? If $y=1$ then $2^y=2\neq 3^2$. – Arthur Feb 27 '18 at 06:32
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Hint: How might multiplying the two given equations help? – Ethan Bolker Mar 04 '18 at 18:32
6 Answers
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Hint
$$2^{3x}=18=2\times 9=2\times 2^y=2^{y+1} \implies y=3x-1$$
Can you finish ?
user577215664
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$$2^{3x} = 18$$
$$2^y = 3^2$$ $\mapsto 2^{y+1} = 9 * 2 =18$
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 $y$
$= \frac{3x + (3x-1)}{6x-1}$ $= \frac{6x-1}{6x-1}$ $=1$
$$\therefore \frac{3x+y}{6x-1} = 1$$
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1Can someone inform me about the error in my answer? I am ready to correct it. Please just don't leave a downvote without any explanation – Aditya Pratap Singh Feb 27 '18 at 07:07
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1i like your answer. You were likely down voted by someone mad that they didn't think of it themselves. – MaximusFastidiousIrreverence Feb 27 '18 at 07:15
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1Welcome to the club of the victims of unjustified downvotes ! $\to +1$. – Claude Leibovici Feb 27 '18 at 07:18
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$3x=\log_218$ and $y=\log_29$.
$3x+y=\log_2(18\times9)$
$6x-1=2\log_218-1=\log_2(18\times18\div2)=\log_2(18\times9)$
$\displaystyle \frac{3x+y}{6x-1}=1$
CY Aries
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By the given $$x=\frac{1}{3}\log_218$$ and $$y=\log_29.$$ Thus, $$\frac{3x+y}{6x-1}=\frac{\log_218+\log_29}{2\log_218-1}=\frac{\log_2(18\cdot9)}{\log_2\frac{18^2}{2}}=\frac{\log_2162}{\log_2162}=1.$$
Michael Rozenberg
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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.
Mark Bennet
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I didn't get this step: $2^{6x-1}=\frac 12 2^{6x}=\frac 12 (2^{3x})^2=\frac 12(18)^2$ – Goendo Feb 27 '18 at 11:39
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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
$$\frac{3x+y}{6x-1} = \frac{3 \cdot\frac{\ln(18)}{3 \cdot \ln(2)}+\frac{\ln(9)}{\ln(2)}}{6 \cdot \frac{\ln(18)}{3 \cdot \ln(2)}-1}=\frac{\frac{\ln(18)+\ln(9)}{\ln(2)}}{\frac{2 \ln(18)-\ln(2)}{\ln(2)}} = \frac{\ln(18) + \ln(9)}{2\cdot \ln(18) - \ln(2)} = \frac{\ln(18 \cdot 9)}{\ln(\frac{18^2}{2})} = \frac{\ln(162)}{\ln(162)} = 1.$$
Bergson
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