Series with Logs

[sick scroll]

Part A and B are quite easy. Can someone help me with part C. I am not sure where to begin.

[wild nexus]

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+close

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[vocal hull]

$\gpterm$

[fathom bluff]

wolfqz

[vocal hull]

@sick scroll

[sick scroll]

I used thaty

I simplified it further to (3^n)/2^(3n-1)

[vocal hull]

wait what

slow down

[sick scroll]

Don't know how to reach to the same statement I have to get it to

[vocal hull]

$a_n = 12 \times \left(\frac{3}{8}\right)^{n-1}$

$a_n = 4 \times 3 \times \frac{3^{n-1}}{8^{n-1}}$

$a_n = 2^2 \times \frac{3^n}{2^{2n-2} \times 2^{n-1}}$

$a_n = \frac{3^n}{2^{3n-5}}$

[fathom bluff]

wolfqz

[vocal hull]

@sick scroll can you continue?

[sick scroll]

wow, i subtracted the powers wwrong on that last step you did

okay let me try

one second

[vocal hull]

oh rip

[vocal hull]

use the property $\log_d \left(\frac{a}{b} \right) = \log_d a - \log_d b, d>0, d \neq 1$

[fathom bluff]

wolfqz

[sick scroll]

log base 2 (An) = log base 2 (3)^n - log base 2 (2)^3n-5

[vocal hull]

yes

[sick scroll]

then the log 2 base 2 cancels, 3n-5 stays

[vocal hull]

$\log_d a^b = b\log_d a$

[fathom bluff]

wolfqz

[sick scroll]

and then we pull the n out from log base 2 (3)^n to n log base 2 (3)

and the statement is complete

thanks a ton

[vocal hull]

np

[sick scroll]

i made a dumb algebra error

thanks

[vocal hull]

u can close the thread now if youre done

[sick scroll]

+close