$\frac{27^{t}}{3^{(t-1)}} = 3 \sqrt{3} $
Can someone please supply a hint to enable me to unlock this? It's fairly entry-level stuff I know and - full disclosure - I'm attempting to assist my daughter with homework, but just cannot see what I'm missing.
1 Answer
You have the correct answer. Here is how I would do this:
First note that
\[\frac{27^t}{3^{t-1}}=\frac{(3^3)^t}{3^{t-1}}=\frac{3^{3t}}{3^{t-1}}=3^{3t-(t-1)}=3^{2t+1}.\]
Also
\[3\sqrt{3}=3\times 3^{\frac{1}{2}}=3^{\frac{3}{2}}.\]
Therefore the equation
\[ \frac{27^t}{3^{t-1}}= 3\sqrt{3}\]
implies
\[ 3^{2t+1}= 3^{\frac{3}{2}}.\]
Hence
\[2t+1=\frac{3}{2} \Rightarrow 2t=\frac{1}{2} \Rightarrow t=\frac{1}{4}.\]
Erdos
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Let me know if you have any questions.
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I may have cracked it. As 27^t is also (3x3x3)^t, 3^(t-1) is also 3^t x 3^-1 and 3 sqrt 3 is also 3 x 3^1/2... the equation does then become: 3 x 3^3t / 3^t = 3 sqrt 3; then 3^2t = sqrt 3, or 3^2t = 3^1/2 The only bit I'm dubious about is then saying that because of this that 2t = 1/2. It does give the correct answer that t = 1/4 though.