Recall that the expression $\log_ba$logba is equivalent to the number $c$c, where $c$c is the power that $b$b must be raised to in order to create $a$a.
For example, we can change $\log_3729$log3729 to the number $6$6 if we recognise that $3^6=729$36=729. This is key to understanding of what a logarithm is.
We can think of a logarithm graph as a curve where, for some base $b$b, the $x$x-axis consists of a set of numbers and the $y$y-axis is the powers of $b$b required to create those numbers. For example, somewhere along the $x$x-axis we will find the number $729$729. The graph of the function $f\left(x\right)=\log_3x$f(x)=log3x will map $729$729 to the number $6$6, simply because $3^6=729$36=729.
Consider $f\left(x\right)=\log_5x$f(x)=log5x. If $x=0.0016$x=0.0016, then $f\left(0.0016\right)=\log_5\left(0.0016\right)$f(0.0016)=log5(0.0016). Before we reach for the calculator, we might recognise that $0.0016=\frac{1}{625}=\frac{1}{5^4}=5^{-4}$0.0016=1625=154=5−4. Since we want the power that $5$5 must be raised to in order to create $0.0016$0.0016, the answer presents itself as $-4$−4. Thus $f\left(0.0016\right)=-4$f(0.0016)=−4.
Base $10$10 logarithms, also known as Briggsian logarithms (after Henry Briggs who compiled them in 1620) or Common logarithms were originally used to multiply and divide numbers. Tables of logs, rather than calculators, were widely available for use by engineers, astronomers, navigators, architects, bankers, teachers, dressmakers, accountants and many other professions. Because they were so common, they became known as common logarithms, and the practise of dropping the base from log expressions soon followed.
So $\log_{10}a$log10a became simply $\log a$loga, and practitioners could recite particular values from memory - $\log2=0.3010$log2=0.3010, $\log3=0.4771$log3=0.4771, so that $\log6=\log\left(2\times3\right)=\log2+\log3=0.7781$log6=log(2×3)=log2+log3=0.7781 etc.
Today, common logarithms are not as special as they were, because we can access logs of other bases quite readily as well, and calculators have eliminated the need to hand calculate products and quotients of numbers. Still, the practise of dropping the base $10$10 part persists in some texts, and so if you see something like $\log10$log10, you can probably assume that the answer is $1$1. Because we would assume that $\log10$log10 means $\log_{10}10$log1010
Study the following five examples carefully.
If $f\left(x\right)=\log x$f(x)=logx, then, assuming base $10$10, $f\left(0.01\right)=\log\left(0.01\right)=\log\left(10^{-2}\right)=-2$f(0.01)=log(0.01)=log(10−2)=−2.
If $f\left(x\right)=\log_2\left(3x-2\right)$f(x)=log2(3x−2), then $f\left(22\right)=\log_2\left(3\times22-2\right)=\log_2\left(64\right)=6$f(22)=log2(3×22−2)=log2(64)=6, since $2^6=64$26=64
If $f\left(x\right)=\log_7\left(2-x\right)$f(x)=log7(2−x), then $f\left(-5\right)=\log_7\left(2--5\right)=\log_7\left(7\right)=1$f(−5)=log7(2−−5)=log7(7)=1. Note here that the argument $\left(2-x\right)$(2−x) must be kept positive. Thus the domain of this function is found by setting $\left(2-x\right)>0$(2−x)>0, from which we can conclude that $x<2$x<2.
If $f\left(x\right)=2\log_{\frac{3}{4}}\left(x-1\right)$f(x)=2log34(x−1), then $f\left(1.5625\right)$f(1.5625) can be evaluated as follows:
$f\left(1.5625\right)$f(1.5625) | $=$= | $2\log_{\frac{3}{4}}\left(1.5625-1\right)$2log34(1.5625−1) |
$=$= | $2\log_{\frac{3}{4}}\left(\frac{9}{16}\right)$2log34(916) | |
$=$= | $2\log_{\frac{3}{4}}\left[\left(\frac{3}{4}\right)^2\right]$2log34[(34)2] | |
$=$= | $2\times2$2×2 | |
$=$= | $4$4 | |
Suppose we know that $f\left(x\right)=\log_2\left(x-1\right)$f(x)=log2(x−1) and that $g\left(x\right)=\log_2\left(x+1\right)$g(x)=log2(x+1). Can we determine the value of $x$x for which the function given by $h\left(x\right)=f\left(x\right)+g\left(x\right)$h(x)=f(x)+g(x) equals $3$3?
To answer this we first note that:
$h\left(x\right)$h(x) | $=$= | $\log_2\left(x-1\right)+\log_2\left(x+1\right)$log2(x−1)+log2(x+1) |
$=$= | $\log_2\left[\left(x-1\right)\left(x+1\right)\right]$log2[(x−1)(x+1)] | |
$=$= | $\log_2\left(x^2-1\right)$log2(x2−1) | |
Then setting $h\left(x\right)=3$h(x)=3, we have that:
$\log_2\left(x^2-1\right)$log2(x2−1) | $=$= | $3$3 |
$x^2-1$x2−1 | $=$= | $8$8 |
$x^2$x2 | $=$= | $9$9 |
$x$x | $=$= | $\pm3$±3 |
Of course the negative solution $-3$−3 must be discarded, because it is outside the domains of the separate functions $f\left(x\right)=\left(x-1\right)$f(x)=(x−1) with $x>1$x>1 and $g\left(x\right)=\left(x+1\right)$g(x)=(x+1) with $x>-1$x>−1. So the correct answer is $x=3$x=3 and we can verify that:
$h\left(3\right)$h(3) | $=$= | $f\left(3\right)+g\left(3\right)$f(3)+g(3) |
$=$= | $\log_2\left(3-1\right)+\log_2\left(3+1\right)$log2(3−1)+log2(3+1) | |
$=$= | $\log_2\left(2\right)+\log_2\left(4\right)$log2(2)+log2(4) | |
$=$= | $1+2$1+2 | |
$=$= | $3$3 | |
Evaluate $\log_50.2$log50.2.
Consider the function $f\left(x\right)=\log_{10}\left(3x+9\right)$f(x)=log10(3x+9).
Evaluate $f\left(4\right)$f(4) to two decimal places.
A function is defined as $f\left(x\right)=\log_2x+\log_2\left(3x-4\right)$f(x)=log2x+log2(3x−4) and another function is defined as $g\left(x\right)=\log_2\left(4x-4\right)$g(x)=log2(4x−4).
Evaluate $f\left(2\right)$f(2).
Evaluate $g\left(2\right)$g(2).
Is $f\left(x\right)=g\left(x\right)$f(x)=g(x)?
Yes
No