Question

Given log 10 = 1 and log 2 = a what can we afirm?
a) log 40 = 1 + 2^a
b) log 40 = 2 + a
c) log 5 = a - 1
d) log 5 = 2 - a

So I proved wrong both a and b:
log 40 => log 4 + log 10 = 2log 2 + 1 => 2a + 1
I don't know what to do with the other options. Help pls!

Answer 1

\[\log(40)=\log(2^2 \cdot 10)=\log(2^2)+\log(10)=2\log(2)+\log(10)=2a+1\]
\[\log(5)=\log(\frac{10}{2})=\log(10)-\log(2)=1-a\]

Answer 2

I'm not getting any of that

Answer 3

i was gonna say that too.. i sat here solving that nonsense for a while..

Answer 4

lol poor bahrom

Answer 5

Me too. Weird... Thx guys

Answer 6

i don't get it :(