Unit 2 Topic 2.12

Logarithmic Function Manipulation

see every function move

3 min readfreeNot practiced

Exponent rules have twins for logarithms. Because a logarithm is an exponent, every rule for combining exponents becomes a rule for combining logarithms. This page states the three properties, the change of base formula, what each one means for a graph, and the domain trap that costs points.

Three properties of logarithms

For a base b>0b > 0, b≠1b \ne 1, and positive numbers xx and yy:

PropertyLogarithm ruleExponent rule behind it
Productlogb⁡(xy)\log_b(xy)=logb⁡x+logb⁡y= \log_b x + \log_b ybm⋅bn=bm+nb^m \cdot b^n = b^{m + n}
Quotientlogb⁡xy\log_b \frac{x}{y}=logb⁡x−logb⁡y= \log_b x - \log_b ybmbn=bm−n\frac{b^m}{b^n} = b^{m - n}
Powerlogb⁡(xn)=nlogb⁡x\log_b(x^n) = n\log_b x(bm)n=bmn(b^m)^n = b^{mn}

Why the product property holds: write x=bmx = b^m and y=bny = b^n, so m=logb⁡xm = \log_b x and n=logb⁡yn = \log_b y. Then xy=bm+nxy = b^{m + n}, so logb⁡(xy)=m+n\log_b(xy) = m + n.

There is no rule for logb⁡(x+y)\log_b(x + y), and logb⁡xy\log_b \frac{x}{y} is not logb⁡xlogb⁡y\frac{\log_b x}{\log_b y}. Those two slips are the most common log errors on the exam.

Expanding and condensing

Worked example 1 (expand). Write log5⁡(25x3)\log_5(25x^3), for x>0x > 0, as a sum of simple terms.

Product: log5⁡25+log5⁡x3\log_5 25 + \log_5 x^3.

Evaluate and use the power property: 2+3log5⁡x2 + 3\log_5 x.

Worked example 2 (condense). Write 2log⁡x+log⁡3−log⁡52\log x + \log 3 - \log 5 as a single logarithm.

Coefficients become exponents first: 2log⁡x=log⁡x22\log x = \log x^2.

Then the sum: log⁡x2+log⁡3\log x^2 + \log 3=log⁡(3x2)= \log(3x^2).

Then the difference: log⁡(3x2)−log⁡5\log(3x^2) - \log 5=log⁡3x25= \log \frac{3x^2}{5}, for x>0x > 0.

"A single logarithm" means exactly one log term. Stopping at log⁡(3x2)−log⁡5\log(3x^2) - \log 5 does not earn the point.

Change of base

Any logarithm can be written through another base aa:

logb⁡x=loga⁡xloga⁡b.\log_b x = \frac{\log_a x}{\log_a b}.

With a calculator, use a=ea = e or a=10a = 10: log7⁡20\log_7 20=ln⁡20ln⁡7= \frac{\ln 20}{\ln 7}≈1.540\approx 1.540. Check the size: 7<20<497 < 20 < 49, so the answer must be between 11 and 22.

The formula also says something about graphs. 1loga⁡b\frac{1}{\log_a b} is a constant, so logb⁡x\log_b x is a constant times loga⁡x\log_a x. For example, log4⁡x\log_4 x=log2⁡xlog2⁡4= \frac{\log_2 x}{\log_2 4}=12log2⁡x= \frac{1}{2}\log_2 x. Every logarithmic function is a vertical dilation of every other one.

What the properties do to graphs

The product property turns a horizontal dilation into a vertical translation. For h(x)=log2⁡(4x)h(x) = \log_2(4x):

h(x)=log2⁡4+log2⁡x=2+log2⁡x.h(x) = \log_2 4 + \log_2 x = 2 + \log_2 x.

So multiplying every input by 44 (a horizontal dilation by a factor of 14\frac{1}{4}) gives the same graph as moving log2⁡x\log_2 x up 22 units. The power property turns a power on the input into a vertical dilation: log2⁡(x3)=3log2⁡x\log_2(x^3) = 3\log_2 x for x>0x > 0.

The domain trap

Each property holds only where every logarithm in it is defined. Look at ln⁡(x2)\ln(x^2) and 2ln⁡x2\ln x.

Worked example 3. Is ln⁡(x2)=2ln⁡x\ln(x^2) = 2\ln x for every x≠0x \ne 0?

At x=−1x = -1: ln⁡((−1)2)=ln⁡1=0\ln((-1)^2) = \ln 1 = 0, but 2ln⁡(−1)2\ln(-1) is undefined.

So they agree only for x>0x > 0. For every x≠0x \ne 0, the true statement is ln⁡(x2)=2ln⁡|x|\ln(x^2) = 2\ln|x|.

When a problem asks for an expression equal to another "for every xx in its domain", check a negative input before you use the power property.

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