Question

I can't seem to solve this :

log(x-3)·log(x)
---------------------- ≤ 0
log(x-4)


I would love a good explanation of the resolution process

Answer 1

Answer:

The given equation is

log(x-3)·log(x)

---------------------- ≤ 0

log(x-4)

As $\log1=0$,

So, we can write the above equation as

log(x-3)·log(x)  ≤ 0 ∧ log(x-4)≥0[ if 1/a≤0, then a≥0]

Now, either log(x-3)≤ 0  ∨ log(x)  ≤ 0  ∨   log(x-4)≥0

Now replacing 0 by $\log1$.

The above equation becomes

log(x-3)≤$\log1$    ∧    log(x)  ≤$\log1$   ∧  log(x-4)≥$\log1$

Cancelling log from both sides of all the three equations

⇒ x-3≤1  ∧ x≤1 ∧  x-4≥1

⇒x≤4  ∧   x≤1     ∧  x≥5

$\frac{\log(x-3)}{\log(x-4)}$ [ if,$x\leq4$this expression has no meaning, so for this expression to exist,$x\geq 5$  ] 

Combining all the three solution , the result of the equation is

  0≤ x≤1 for $\log x \text {and } x> 4 for \frac{\log(x-3)}{\log(x-4)}$

Why i have written this because the whole expression should be less than zero, so when 0≤ x≤1 ,$\log x$ becomes negative and when x>4 the expression $\frac{\log(x-3)}{\log(x-4)}$ will be positive.

Answer 2

$D:x-3>0 \wedge x>0 \wedge x-4>0 \wedge \log(x-4)\not=0\\D:x>3 \wedge x>0 \wedge x>4 \wedge x-4 \not=1\\D:x>4 \wedge x\not=5\\D:x\in(4,5)\cup(5,\infty)\\\\\dfrac{\log(x-3)\cdot \log x}{\log(x-4)}\leq0\\\\\log(x-3)\cdot \log x \cdot \log (x-4)\leq0\\\\x_0=4 \vee x_0=1 \vee x_0=5\\\\x\in(-\infty,1\rangle\cup\langle4,5\rangle\\\\\\x\in(-\infty,1\rangle\cup\langle4,5\rangle \wedge x\in(4,5)\cup(5,\infty)\\\\\boxed{\boxed{x\in(4,5)}}$