Logarithms

Hong Kong

Stage 4 - Stage 5

Lesson

The quantity $\log_ax$`l``o``g``a``x` is defined for positive values of $x$`x` only. More generally, if we replace $x$`x` with some expression of $x$`x`, for instance $3x-4$3`x`−4, then $\log_a\left(3x-4\right)$`l``o``g``a`(3`x`−4) is defined if and only if $3x-4>0$3`x`−4>0.

If we directly solve an inequality involving logs, our solution might look like the following:

$x>c,x\ge c,x`x`>`c`,`x`≥`c`,`x`<`c`,`x`≤`c`

where $c$`c` is some constant.

It may be tempting to stop and finish here. However, we want to check whether the solution guarantees that each log is defined. In our case, that means making sure that the solution, say $x>c$`x`>`c`, also means $3x-4>0$3`x`−4>0. Let's explore this using the following example.

Suppose we wish to solve the inequality given by $\log_2\left(3x-4\right)<5$`l``o``g`2(3`x`−4)<5.

We might proceed as follows:

$\log_2\left(3x-4\right)$log2(3x−4) |
$<$< | $5$5 | (Given) |

$2^{\log_2\left(3x-4\right)}$2log2(3x−4) |
$<$< | $2^5$25$\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }$ | (If $aa<b then $2^a<2^b$2a<2b) |

$3x-4$3x−4 |
$<$< | $2^5$25 | ($a^{\log_ab}=b$alogab=b) |

$3x-4$3x−4 |
$<$< | $32$32 | (Simplifying the power) |

$3x$3x |
$<$< | $36$36 | (Adding $4$4 to both sides) |

$x$x |
$<$< | $12$12 | (Dividing both sides by $3$3) |

One important thing to note here is that whenever $a`a`<`b`, we have $2^a<2^b$2`a`<2`b`. This is because the graph of $y=2^x$`y`=2`x` is a strictly increasing function. In fact this is true for any graph $y=k^x$`y`=`k``x` (so long as $k>1$`k`>1).

Looking at our final line of working, we might then incorrectly decide that the solution set is all $x$`x` such that $x<12$`x`<12 as shown here on the number line.

However, there are values of $x$`x` here that will violate the fact that the expression inside the log must remain positive.

So our next step is to solve when the expression inside the log is greater than zero. In other words, we need to solve the inequality $3x-4>0$3`x`−4>0 and the solution to that is $x>\frac{4}{3}$`x`>43, shown graphically here:

Putting these two things together, we must adjust our first solution so that it becomes a subset of the condition $x>\frac{4}{3}$`x`>43. In other words, we seek all values of $x$`x` that are less than $12$12, but also greater than $\frac{4}{3}$43. Visually, this means the values of $x$`x` within the blue line below:

The correct solution is then given by $\frac{4}{3}`x`<12.

In practice, it's not necessary to draw diagrams, but it does nicely complement the process.

When we're asked, for example, to solve something like $\log_2x\le3$`l``o``g`2`x`≤3, we should keep in mind that the solution we arrive at algebraically (in this case $x\le9$`x`≤9) does not take into account that the expression inside the log is positive: $x>0$`x`>0. In other words, for this example, our solution becomes $0`x`≤9.

There are two interesting properties that will help solve an inequality with logs. These come from the fact that both exponential curves and log curves are strictly increasing functions.

Consider the function $f\left(x\right)=6\log_44x-12$`f`(`x`)=6`l``o``g`44`x`−12.

Solve $f\left(x\right)=0$

`f`(`x`)=0.Solve $f\left(x\right)>0$

`f`(`x`)>0.

Consider the inequality $\log_5x<-2$`l``o``g`5`x`<−2.

State the domain of $\log_5x$

`l``o``g`5`x`.Hence solve the inequality $\log_5x<-2$

`l``o``g`5`x`<−2.

Consider the inequality $\log_7\left(x-5\right)\le\log_7\left(3x-9\right)$`l``o``g`7(`x`−5)≤`l``o``g`7(3`x`−9).

State the domain of $\log_7\left(x-5\right)$

`l``o``g`7(`x`−5).State the domain of $\log_7\left(3x-9\right)$

`l``o``g`7(3`x`−9).Hence solve the inequality $\log_7\left(x-5\right)\le\log_7\left(3x-9\right)$

`l``o``g`7(`x`−5)≤`l``o``g`7(3`x`−9).