Question

help please Finding the domain

Answer 1

\[9-7\log_{2} \ (\frac{ x }{ 7 }-6)\]

Answer 2

You cannot take the log of a number less than or equal to 0.
So log(A) is defined only for A > 0.
You can use that to find the domain.

Answer 3

so would it be \[(\frac{ 6 }{ 7 }-\infty)\]

Answer 4

No. When taking the log of (x/7 - 6) we have to make sure (x/7 - 6) > 0
Solve the inequality: (x/7 - 6) > 0

Answer 5

\[x >42\]

Answer 6

Yes. All values of x > 42 is fine.
How would you express that in interval form?

Answer 7

\[(42,\infty)\]

Answer 8

Yes. That is it!

Answer 9

thank you so much

Answer 10

you are welcome.

Answer 11

one more question could i approach this problem \[e^{x ^{2}}= e ^{16x}*\frac{ 1 }{ e ^{60} }\]

Answer 12

how^

Answer 13

\[\Large e^{x ^{2}}= e ^{16x}*\frac{ 1 }{ e ^{60} }\]

Take natural log on both sides and solve for x.

Answer 14

\[\Large \log(a)^m = m * \log(a)\]

Answer 15

\[\Large \log(A*B) = \log(A) + \log(B)\]
\[\Large \frac{ 1 }{ a^m } = a^{-m}\]

Answer 16

\[x ^{2}=x*\log_{16} * e ^{-60}\]

Answer 17

?

Answer 18

\[\Large \ln(e^{x ^{2}})= \ln(e ^{16x}*\frac{ 1 }{ e ^{60} }) = \ln(e ^{16x}) + \ln(\frac{ 1 }{ e ^{60} })\]

Answer 19

\[\Large x^2 = 16x + \ln(e^{-60}) = 16x - 60\]

Answer 20

so in there I can use as \[x ^{2}-16x+60= (x-10)(x-6)= answers would be X=6,10\]

Answer 21

Yes.

Answer 22

thanks

Answer 23

yw.