Can someone please help me? log_2 5 - log_2 3 30,000 = (15,500)(1.18)^x
@sanaybrown you mind helping me out?
@MelodicSymphony do you mind helping me out?
What's your question?
log_2 5 - log_2 3 and 30,000 = (15,500)(1.18)^x
What is it asking?
All you gave is an expression that means nothing, \[\log_2 5 - \log_2 3 \] what do you want to do with this?
I think they want it solved as in what does that expression equal
It's not a function you can't solve it
for the first one its asking to choose the single logarithmic expression that is equivalent to the one shown. Log2 Log_2 Log_2 5/3 Log_2 3/5
Oh ok, so that makes sense now. When you are given such an expression you use the quotient rule, apply the following property. \[\log_a \frac{ x }{ y } = \log_a x- \log_a y\]
I think it's C but i'm not sure if i'm right
Yup.
okay good. how about the sec question?
What is it asking/
Suppose the population in a town has an average growth rate of 18% . If there are 15,500 people now living in town, in about how many years will the population grow to 30,000? Use the equation 30,000 = (15,500)(1.18)x and round the value of x to the nearest year.
the answers are 4 5 6 7
I went with 6, im I correct?
How did you get it?
You know if I told you the answer, that would not help you in any way, so if you can show me your work I can tell you if you did it correctly or not.
I solved for x and got x=6.04176898ln(3000)
Mhm
\[30000 = (15500)(1.18)^x\] \[\frac{ 30,000 }{ 15,500 } = (1.18)^x\]your first step correct?
\[\implies \frac{ 60 }{ 31 } = (1.18)^x \implies \ln \left( \frac{ 60 }{ 31 } \right) = \ln(1.18)^x\] what's next?
I tried to do it again and got x≈3.9897266
Yeah :)
That looks good, just round it up
so if i round it up it will = to 4 years
\[\huge \ln \left( \frac{ 60 }{ 31 } \right) = x \ln(1.18) \implies x = \left( \frac{ \ln \frac{ 60 }{ 31 } }{ \ln(1.18) } \right)\]
do u mind checking more answers out? I want to know if my work is good
\[\log_a (x^p) = p \log_a\] (power property)
Open up a new question :)
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