Question

solve for x. log(x)-log(x+9)=1

Answer 1

Use the fact that log(a) - log(b) = log(a/b) for your first step

Answer 2

Yeah i got \[\log(x/x+9)=1\] what do i do from here

Answer 3

Do you know the rule for adding and subtracting logs? Two hundred years ago engineers and scientists made tables of logs of every number because large number multiplications are turned into addition with logs.
Say you want to know what 137890*98900 is equal to:
If you have a table of logs, you will know that log(137890)+log(98900) = log(137890*98900) = 10.133
and 10^10.133 = 1.358*10^10

Answer 4

So, from where you are, exponentiate both sides by 10.

Answer 5

what does that mean ?

Answer 6

if you know that log(pi) = x, then use each side of the equality as exponents to make a new equation.
On one side you'll have
\[10^{\log(\pi)}\]
and on the other side
\[10^{x}\]

the \[10^{\log(\pi)}\] simplifies to pi

Answer 7

\(\bf log(x)-log(x+9)=1\implies log\left(\cfrac{x}{x+9}\right)=1\implies log_{10}\left(\cfrac{x}{x+9}\right)=1\\ \quad \\
\textit{log cancellation rule of }\quad a^{log_ax}=x\qquad thus\\ \quad \\
\Large log_{10}\left(\cfrac{x}{x+9}\right)=1\implies 10^{log_{10}\left(\frac{x}{x+9}\right)}=10^1\\ \quad \\\implies \cfrac{x}{x+9}=10\)

Answer 8

hmmm shoot got a bit truncated

Answer 9

\(\bf log(x)-log(x+9)=1\implies log\left(\cfrac{x}{x+9}\right)=1\\ \quad \\\implies log_{10}\left(\cfrac{x}{x+9}\right)=1\\ \quad \\
\textit{log cancellation rule of }\quad a^{log_ax}=x\qquad thus\\ \quad \\
\large log_{10}\left(\cfrac{x}{x+9}\right)=1\implies 10^{log_{10}\left(\frac{x}{x+9}\right)}=10^1\\ \quad \\
\implies \cfrac{x}{x+9}=10\)

Answer 10

Thank both of you for your help. :)

Answer 11

Hey @jdoe0001, how do you get your equation fonts to be bold and use the arrow?

Answer 12

\bf <--- turns on BoldFace \implies -> \(\implies\)

Answer 13

@brendafn happy to help.

Answer 14

Thanks

Answer 15

@Matt.Mawson http://www.onemathematicalcat.org/MathJaxDocumentation/TeXSyntax.htm

Answer 16

:)

Answer 17

@jdoe0001 do you know how i would check, because when i plug in 10 to the orginal equation i get -0.27875