Question

Logarithm probems... Please help!?

Answer 1

1. Solve log(3x + 2) = 3

Answer 2

x = 302.67 :)

Answer 3

3x+2=10^3
follow that

Answer 4

what is that in fraction form? Its multiple choice and they are all in fraction form

Answer 5

ok.. first you ca apply the follow equation:\[\log _{a}b=c--->a ^{c}=b\].. later you have the equation: \[3x + 2=10^{3}\] finally, you have: \[x=\frac{ 10^{3} -2}{ 3}\]

Answer 6

a. \[\frac{ 1000 }{ 3 }\]
b.\[\frac{ 998 }{ 3 }\]
c. \[998\]
d. \[\frac{ 1 }{ 3 }\]

These are the choices

Answer 7

So would it be B ?

Answer 8

then, option B :D

Answer 9

thanks, could you help me with two more please ?

Answer 10

off course :D

Answer 11

2. log(x + 9) - log x = 3

3. \[5 \log _{b}y + 6 \log _{b}x\]

Answer 12

For number 2. The options are a. 0.0090, b. 0.3103, c. 3.2222, d. 111

Answer 13

log(x+9)-logx=log[(x+9)/x]=3 follow that.

Answer 14

for the second exercise, you have \[\log (x+9)-logx=3-->\log(\frac{ x+9 }{3})=3\].. then, \[10^{3}=\frac{ x+9 }{3}-->x=3(10^{3})-9\]

Answer 15

I got 2991...

Answer 16

yeah, thet's the answer :D

Answer 17

But that isn't one of my choices

Answer 18

D: there are an error sorry

Answer 19

which one would i choose then?

Answer 20

hen the equation is \[\log(x+9)-logx=3-->\log(\frac{ x+9 }{x })=3\].. then, \[10^{3}=\frac{ x+9 }{ x }-->10000x-x=9-->x=\frac{ 9 }{999}\]

Answer 21

option "a" for the second exercise

Answer 22

Okay I see now. What about number three?

Answer 23

Choices are \[a. \log _{b}(yx ^{5+6})\]
\[b. (5 + 6) \log _{b} (y + x)\]
\[ c. \log _{b} (y ^{5} + x ^{6})\]
\[d. \log _{b} (y ^{5}x ^{6})\]

Answer 24

well the third exercise, you have to know that:\[A \times \log _{b}c=\log _{b}c ^{A}\].. following this equation, we have \[\log _{b}y ^{5} + \log _{b}y ^{6}\] and for logs thet same base, you can:\[\log _{b}a+\log _{b}c=\log _{b}(a \times c)\] then in our exercise, the answer is \[\log _{b} (y ^{5}x ^{6})\] option D :D

Answer 25

OHH Okay, Thank you so much for your help! (:

Answer 26

:)