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equivalent equation to y=2.5(2)t/5 show work

 Jun 5, 2014

Best Answer 

 #3
avatar+118667 
+5

Alan's answer is excellent but I would not have thought to do it that way.

also I often refer to this page when I do log questions.

http://en.wikipedia.org/wiki/List_of_logarithmic_identities

Here is  a more mainstream solution.

$$\begin{array}{rll}
y&=&2.5\times 2^{t/5}\\\\
\frac{y}{2.5}&=&2^{t/5}\\\\
log\left(\frac{y}{2.5}\right)&=&log(2^{t/5})\\\\
log\left(\frac{y}{2.5}\right)&=&(t/5)log(2)\\\\
\frac{log\left(\frac{y}{2.5})}{log(2)}\right)&=&t/5\\\\
\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)&=&t\\\\
t&=&\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)\\\\
t&=&\frac{5log\left(0.4y)}{log(2)}\right)\\\\
\end{array}$$

I'd stop here but i want to check it is the same as Alan's solution.

I know that 

$$\frac{log_{10}A}{log_{10}B}=Log_BA$$

so

$$t=5log_2(y/2.5)\\

t=5log_2(0.4y)$$

They are the same.    

 Jun 6, 2014
 #1
avatar+33659 
+5

I assume that by "equivalent equation" you mean rearrange to get t = ...

Take log to the base 2 of both sides

log2y = log2(2.5*2t/5)

log2y = log2(2.5) + log2(2t/5)

log2y = log2(2.5) + t/5

t/5 = log2y - log2(2.5)

t/5 = log2(y/2.5)

t = 5log2(y/2.5)

 Jun 5, 2014
 #2
avatar+154 
0

Thanks for your help!

 Jun 5, 2014
 #3
avatar+118667 
+5
Best Answer

Alan's answer is excellent but I would not have thought to do it that way.

also I often refer to this page when I do log questions.

http://en.wikipedia.org/wiki/List_of_logarithmic_identities

Here is  a more mainstream solution.

$$\begin{array}{rll}
y&=&2.5\times 2^{t/5}\\\\
\frac{y}{2.5}&=&2^{t/5}\\\\
log\left(\frac{y}{2.5}\right)&=&log(2^{t/5})\\\\
log\left(\frac{y}{2.5}\right)&=&(t/5)log(2)\\\\
\frac{log\left(\frac{y}{2.5})}{log(2)}\right)&=&t/5\\\\
\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)&=&t\\\\
t&=&\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)\\\\
t&=&\frac{5log\left(0.4y)}{log(2)}\right)\\\\
\end{array}$$

I'd stop here but i want to check it is the same as Alan's solution.

I know that 

$$\frac{log_{10}A}{log_{10}B}=Log_BA$$

so

$$t=5log_2(y/2.5)\\

t=5log_2(0.4y)$$

They are the same.    

Melody Jun 6, 2014

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