Question

write the expression as a single logarithm

1/3 log6 z + 7 log6 y - log6 w

Answer 1

Using the properties\[log_{b}(a)+log_b(c)\Leftrightarrow log_b(a*c)\]\[log_{b}(a)+log_b(c)\Leftrightarrow log_b\left(\frac{a}{c}\right)\]\[c*log_b(a)\Leftrightarrow log_b(a^c)\]Then\[\frac{1}{3}log_6(z)+7log_6(y)-log_6(w)\]\[log_6(z^{\frac{1}{3}})+log_6(y^7)-log_6(w)\]\[log_6\left(\frac{z^{\frac{1}{3}}*y^7}{w}\right)\]

Answer 2

so do I work that out?

Answer 3

or is that my answer

Answer 4

Yes

Answer 5

It's better to you doing it step by step like i did.

Answer 6

im so sorry I'm so confused. i don't understand how to solve that last one

Answer 7

log6 (z1/3*y^7/2)

Answer 8

For example: \[log(a)+3log(b)-2log(c)\]Using the property of exponential\[log(a)+log(b^3)-log(c^2)\]Using the signal property (for Plus)\[log(a*b^3)-log(c^2)\]Using the signal property (for Minus)\[log\left(\frac{a*b^3}{c^2}\right)\]

Answer 9

so the answer is log6(z^1/3 * y^7/w)

Answer 10

Yes ;)

Answer 11

We have a single logarithm ;)

Answer 12

is it z^1/3 or just z 1/3

Answer 13

thanks sorry I got so confused for a sec...thanks so much

Answer 14

\[z^{\frac{1}{3}}\]

Answer 15

can I ask you another question please??

Answer 16

Sure, did you understand my first answer now?!

Answer 17

yes...took me a second but I got it

Answer 18

graph the function g(x)=2^x+2 and give its domain and range using interval notation

Answer 19

It's \[g(x)=2^{x+2}~~~~or~~~~g(x)=2^x+2\]

Answer 20

oh haha the first one

Answer 21

sorry

Answer 22

Domain:\[D=(-\infty,+\infty)=\mathbb{R}\]
Range:\[R=(0,+\infty)=\mathbb{R}^{*-}\]