QUESTION 1
The given logarithmic expression is

We rewrite
in the index form to base 4.
This implies that;

We now apply the power rule:
.

Recall that logarithm of the base is 1.


QUESTION 2
The given logarithm is;
![\log_2(\sqrt[5]{32})](https://tex.z-dn.net/?f=%5Clog_2%28%5Csqrt%5B5%5D%7B32%7D%29)
![\log_2(\sqrt[5]{2^5})](https://tex.z-dn.net/?f=%5Clog_2%28%5Csqrt%5B5%5D%7B2%5E5%7D%29)
This is the same as;



Answer: v = - 9
Step-by-step explanation:
Since two points on the line are given, we use formula for slope.
Let (a,b) and (c,d) be two points on the line, the slope of the line is given by
m=
.
Hence, using this we have
-4=
Thus, we have
-4×(-3)=3-v⇒12=3-v⇒12-3=-v⇒v=-9
1) 40-25= 15
2) 15/3=5
3) each gift costs $5
Then to check:
1) 5x3=15
2) 25+15=40
Therefore p=5
Since 34 only goes into 7 four times the answer is 28 with a remainder of 6:) so basically you would could by your 7s so 7, 14, 21, 28, 35 and since 35 is too much we would have to go with 28. so then we subtract 28 from 34 and that gives us 6 left over. hoped this helped:))
Answer:
x = log 7
Step-by-step explanation:
10^x=7
Take the log base 10 on each side
log10 (10^x)=log10 (7)
We dont need to write the 10, it is implied
log(10^x)=log (7)
The exponent goes out front since log a^b = b log a
x log 10 = log 7
log 10 = 1 since the base is 10 log10(10) =1
x = log 7