Question

How do I find the solution of log(x - 3)30.001 = -3 With (x - 3) being the base?

Answer 1

\[\log_{x-3} .001 = -3\]

Answer 2

Like that actually ^

Answer 3

Well raise everything like so:
\[(x-3)^{\log_{x-3}(10^{-3})} = (x-3)^{-3} \implies 10^{-3} = (x-3)^{-3} \]
\[\implies 10 = x-3 \implies x = 7\]

Answer 4

use the rule \[\Large \log _{a} b = c\]then
\[\Large a^ c = b\]
so
\[\Large \log_{x-3} .001 = -3\]
\[\Large (x-3)^{-3} = 0.001\]

Answer 5

Yeah I would get what agent0smith has, but I would get stuck there. I don't quite understand how you did that though, malevolence19.

Answer 6

Well .001 = 10^(-3) or 1/1000 as 1 is in the thousandths place...

Answer 7

And if you have (f(x))^n = (g(x))^n implies f(x) = g(x)

Answer 8

Oh okay I get it now. Basically you raised both sides by the base of the log which allowed you to have same bases and thus work on only the exponents. From there you were able to cancel out the ^-3 and do easy math from there. Right?

Answer 9

Just want to make sure I understood correctly.

Answer 10

I mean, you aren't 'cancelling' the ^-3 per se. You really have:
\[(x-3)^{-3} = 10^{-3} \rightarrow \frac{1}{(x-3)^3} =\frac{1}{10^3} \rightarrow \frac{1}{x-3} = \frac{1}{10} \implies x-3 = 10\]
But I saved you all the 'justification'.

Answer 11

Ah, so I did that wrong in my head lolz, its 13 not 7 xD

Answer 12

10+3 =/= 7
10+3 = 13

Answer 13

Ohhh okay. Thank you very much!
So if I were to do \[\log_{x + 5}.001 = -3 \] it would be x = 5?

Answer 14

Yup because:
\[\log_{5+5}(10^{-3})=\log_{10}(10^{-3})=-3\]
As:
\[\log_a (a^b) = b\]
So this one you can "eyeball" the correct answer because you know the base must be 10 giving: x+5 = 10 implies x = -5.

Answer 15

Alrighty, thanks for the help :) Really appreciate it!

Answer 16

You can also put both sides to the power of -1/3 and solve from there:

\[\Large ((x-3)^{-3})^{-\frac{ 1 }{ 3 } }= (0.001)^{-\frac{ 1 }{ 3 }}\]

You can do that for any power, you put both sides to the power of 1/a:
\[\Large ((x-3)^{-a})^{-\frac{ 1 }{ a } }= (0.001)^{-\frac{ 1 }{ a }}\]

When you have something to the power of a, you can "undo" it by putting both sides to the power of the reciprocal, which is 1/a.