Question

`log_(8 x^(2) -23x+15)(2x-2) \leq 0`

Answer 1

`\log_(8x^2-23x+15) (2x-2) \leq 0`

The domain:
The number of which the logarithm is taken must be greater than 0.

`2x-2 \ \textgreater \ 0 \\ 2x\ \textgreater \ 2 \\ x\ \textgreater \ 1 \\ x \in (1, +\infty)`

The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:

`8x^2-23x+15\ \textgreater \ 0 \\ 8x^2-8x-15x+15\ \textgreater \ 0 \\ 8x(x-1)-15(x-1)\ \textgreater \ 0 \\ (8x-15)(x-1)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ 8x-15=0 \ \lor \ x-1=0 \\ 8x=15 \ \lor \ x=1 \\ x=(15)/(8) \\ x=1 (7)/(8) \\ \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola op} \hbox{ens upwards} \\ \hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros} \\ \\ x \in (-\infty, 1) \cup (1 (7)/(8), +\infty)`

*not equal to 1:

`8x^2-23x+15 \\ot= 1 \\ 8x^2-23x+14 \\ot= 0 \\ 8x^2-16x-7x+14 \\ot= 0 \\ 8x(x-2)-7(x-2) \\ot= 0 \\ (8x-7)(x-2) \\ot= 0 \\ 8x-7 \\ot=0 \ \land \ x-2 \\ot= 0 \\ 8x \\ot= 7 \ \land \ x \\ot= 2 \\ x \\ot= (7)/(8) \\ x \\otin \{(7)/(8), 2 \}`

Sum up all the domain restrictions:

`x \in (1, +\infty) \ \land \ x \in (-\infty, 1) \cup (1 (7)/(8), +\infty) \ \land \ x \\otin \{ (7)/(8), 2 \} \\ \Downarrow \\ x \in (1 (7)/(8), 2) \cup (2, +\infty)`

The solution:

`\log_(8x^2-23x+15) (2x-2) \leq 0 \\ \\ \overline{\hbox{convert 0 to the logarithm to base } 8x^2-23x+15} \\ \Downarrow \\ \underline{(8x^2-23x+15)^0=1 \hbox{ so } 0=\log_(8x^2-23x+15) 1 \ \ \ \ \ \ \ } \\ \\ \log_(8x^2-23x+15) (2x-2) \leq \log_(8x^2-23x+15) 1`

Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.

* if the base is less than 1:

`8x^2-23x+15 \ \textless \ 1 \\ 8x^2-23x+14 \ \textless \ 0 \\ \\ \hbox{the zeros have already been calculated: they are } x=(7)/(8) \hbox{ and } x=2 \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\ \hbox{the values less than 0 are between the zeros} \\ \\ x \in ((7)/(8), 2) \\ \\ \hbox{including the domain:} \\ x \in ((7)/(8), 2) \ \land \ x \in (1 (7)/(8), 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (1 (7)/(8) , 2)`

The inequality:

`\log_(8x^2-23x+15) (2x-2) \leq \log_(8x^2-23x+15) 1 \ \ \ \ \ \ \ |\hbox{flip the sign} \\ 2x-2 \geq 1 \\ 2x \geq 3 \\ x \geq (3)/(2) \\ x \geq 1 (1)/(2) \\ x \in [1 (1)/(2), +\infty) \\ \\ \hbox{including the condition that the base is less than 1:} \\ x \in [1 (1)/(2), +\infty) \ \land \x \in (1 (7)/(8) , 2) \\ \Downarrow \\ x \in (1 (7)/(8), 2)`

* if the base is greater than 1:

`8x^2-23x+15 \ \textgreater \ 1 \\ 8x^2-23x+14 \ \textgreater \ 0 \\ \\ \hbox{the zeros have already been calculated: they are } x=(7)/(8) \hbox{ and } x=2 \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\ \hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros}`


`x \in (-\infty, (7)/(8)) \cup (2, +\infty) \\ \\ \hbox{including the domain:} \\ x \in (-\infty, (7)/(8)) \cup (2, +\infty) \ \land \ x \in (1 (7)/(8), 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (2, \infty)`

The inequality:

`\log_(8x^2-23x+15) (2x-2) \leq \log_(8x^2-23x+15) 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq (3)/(2) \\ x \leq 1 (1)/(2) \\ x \in (-\infty, 1 (1)/(2)] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 (1)/(2)] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset`

Sum up both solutions:

`x \in (1 (7)/(8), 2) \ \lor \ x \in \emptyset \\ \Downarrow \\ x \in (1 (7)/(8), 2)`

The final answer is:

`\boxed{x \in (1 (7)/(8), 2)}`