Question

will fan and medal first to answer.
solve the equation:
log(x+7)-logx=3

Answer 1

Please review the basic rules of logs:
log a + log b = log ab
log a - long b = log (a/b)
log a^b = b log a

Answer 2

Which of those three rules applies to the given question?

Answer 3

log(x+7)-log(3)=3
log((x+7)/3)=3
{x+7}/3=10^3
x+7=3000
x=2993

Answer 4

\[\log~m-\log~n=\log \frac{ m }{n }\]
\[loga=b,10^{b}=a\]

Answer 5

@mathmale I have to actually solve for x. That's how to write it as one logarithm

Answer 6

Yes, of course. Still, one of these three rules of logs applies to the left side of your math problem. Take another look, please.

Answer 7

Loga-logb=log a/b

Answer 8

Yes. So, log (x+7) - log (3) = log (what fraction?)

Answer 9

Why is your second number log(3)?

Answer 10

it will be log(x+7)-log(x)=log([x+7]/[x])
as loga-logb=log(a//b)

Answer 11

@rishabhjaiswal225499 As one logarithm yes. I have to solve for x though. every time I try it cancels out

Answer 12

i have solved it above you may have a look at it x is not getting cancled

Answer 13

You've put it into one logarithm. I need the value of x. Here are my choices.
A. 3.5
B.142.7
C.0.0070
D.0.0707

Answer 14

Yes. So, log (x+7) - log (3) = log (what fraction? It's important that you answer this.
If log a - log b = log a/b, then log (x-7) - log (3) =
\[\log \frac{ x+7 }{ 3 }=3.\]
Think: How could you solve this for x?
What function is the inverse of y=log x? (That's a hint.)

Answer 15

The second value isn't log(3) its logx

Answer 16

well than maybe base is e and not 10 if its so than
log([x+7]/3)=3
so (x+7)/3=e^3 where e=2.7
so x+7=20.08
x=14.08 i dont know but its crt i am sure

Answer 17

Here's how I'm working it out and you can tell me what's wrong with it.
log(x+7)-logx=3
(x+7)-x=10^3
(x+7)-x=1000
-7 -7
x-x=993
canceled out

Answer 18

@suckerofmath : Thanks for pointing out my mistake. Can you guide rishabh through the problem solution?

Answer 19

you know what if loga - logb =x
than we cannot write a-b =10^x
we have to write log(a/b)=x
so a/b=10^x
and not a-b=10^x your second step is wrong @suckerofmath

Answer 20

the 10^3 is how you get rid of the log because you put everything to the power e and it makes the right side of the equation 10^3
@rishabhjaiswal225499

Answer 21

u can use this \[\log_{a} b=x\]
so\[a ^{x}=b\]

Answer 22

so \[\log_{10} ((x+7)/3)=3 so 10^3=((x+7)/3)\]

Answer 23

@rishabhjaiswal225499 That still doesn't give the value of x

Answer 24

\[\log \frac{ x+7 }{ x }=3\].
This is the common log, so the base is 10.

Write 10 = 10
and then let log (x+7)/x become the exponent of the first base 10 and 3 the exponent of the second:
\[10^{\log \frac{ x+7 }{ 7 }}=10^{3.}\]
Can you now solve the problem?

Answer 25

Now I'm really confused.