Question

if r>0 and r^t=6.25r^t+2, then r=

Answer 1

r= unreal no.

Answer 2

Re-arrange, factor out r^t and then do some more stuff that I'll show:\[\bf r^t-6.25r^t-2 = 0 \implies r^t(1-6.25)-2=0 \implies -5.25r^t-2=0\]Now isolate r^t from the rest of the equation :\[\bf r^t=\frac{ 2 }{ -5.25 }=-\frac{ 8 }{ 21 } \implies r^t + \frac{ 8 }{ 21 }=0\]You can see that if I were to re-arrange and take the natural logarithm I would'nt be able to. Hence we've got no solutions within the realm of Real numbers, considering that r > 0.

Answer 3

Also r > 0 suggests that 'r' is positive and the exponential graph of a positive number to the power 't' would never be 0 hence the sum of r^t and 8/21 such that r > 0 can never be 0 hence there exist no solutions for 'r'.

Answer 4

Unless the question is \(\bf r^t=6.25r^{t+2}\), then we could probably work something out.

@abdullahnasir

Answer 5

If the above is the case, then we work out 'r' by taking the natural logarithm of both sides and re-arranging:\[\bf \ln(r^t)=\ln(6.25r^{t+2}) \implies tln(r)=\ln(6.25)+(t+2)\ln(r)\]Re-arranging and factoring gets us the following:\[\bf tln(r)-(t+2)\ln(r)=\ln(6.25) \implies \ln(r)(t-(t+2))=\ln(6.25)\]\[\bf \implies -2\ln(r) =\ln(6.25) \implies \ln(r)=\ln(6.25)^{-\frac{ 1 }{ 2 }}\]\[\bf \implies r = \frac{ 1 }{ \sqrt{6.25} }=\frac{ 1 }{ \frac{ 5 }{ 2 } }=\frac{ 2 }{ 5 }\]

@abdullahnasir

Answer 6

\[\frac{ r ^{t}}{ r ^{t+2}}=6.25\]
\[r^{2}=\frac{ 1 }{ 6.25 }\]
hence the above answer is correct on the basis if the question is as mentioned.