Introduction to Logarithms: Additional Practice (Set 1)

Work from the inside out. The inner logarithm asks: three to the what gives eighty-one?

$3^4 = 81 \quad \Longrightarrow \quad \log_3 81 = 4$

The expression is now $\log_2 4$, and $2^2 = 4$.

$\log_2 4 = 2$

Switch to exponential form. The statement says $x$ raised to the power $\tfrac{3}{2}$ gives $125$.

$x^{3/2} = 125$

Raise both sides to the reciprocal power $\tfrac{2}{3}$.

$x = 125^{2/3} = \left(125^{1/3}\right)^2 = 5^2 = 25$

Check: $25^{3/2} = \left(25^{1/2}\right)^3 = 5^3 = 125$, so $\log_{25} 125 = \tfrac{3}{2}$.

Write the base $27$ as $3^3$ so the cancellation law can act on the base $3$.

$27^{\log_3 2} = \left(3^3\right)^{\log_3 2} = \left(3^{\log_3 2}\right)^3$

The inner expression collapses, since $3^{\log_3 2} = 2$ by the definition of the logarithm.

$\left(3^{\log_3 2}\right)^3 = 2^3 = 8$

Demand a positive argument and solve the quadratic inequality.

$x^2 - 5x + 6 > 0 \quad \Longrightarrow \quad (x - 2)(x - 3) > 0$

The product of the two factors is positive when both are negative ($x < 2$) or both are positive ($x > 3$).

$x < 2 \quad \text{or} \quad x > 3$

On the interval $2 < x < 3$ the parabola dips below the axis, so the argument is negative there and the logarithm does not exist.

Switch to exponential form: the argument equals $5^1$.

$x^2 - 4x = 5 \quad \Longrightarrow \quad x^2 - 4x - 5 = 0 \quad \Longrightarrow \quad (x - 5)(x + 1) = 0$

So $x = 5$ or $x = -1$. The domain rule restricts the argument, not $x$, so test the argument in each case.

For $x = 5$ it is $25 - 20 = 5 > 0$, and for $x = -1$ it is $1 + 4 = 5 > 0$. Both are positive, so both values solve the equation.

Let $y = \log_{1/5} 125$, so $\left(\tfrac{1}{5}\right)^y = 125$. Write both sides as powers of $5$, using $\tfrac{1}{5} = 5^{-1}$.

$5^{-y} = 5^3 \quad \Longrightarrow \quad -y = 3 \quad \Longrightarrow \quad y = -3$

Check: $\left(\tfrac{1}{5}\right)^{-3} = 5^3 = 125$. The base is less than $1$, so the logarithm of a number greater than $1$ comes out negative.

The point says $\log_b 81 = 4$, which in exponential form says the base raised to the fourth power gives $81$.

$b^4 = 81$

The real roots are $b = 3$ and $b = -3$, but a logarithm base must be positive, so $b = 3$.

Check: $3^4 = 81$, so $\log_3 81 = 4$ and the curve does pass through $(81,\ 4)$.

The asymptote of a logarithm sits where its argument reaches $0$. Adding $2$ shifts the curve up and moves no vertical line.

$x - 7 = 0 \quad \Longrightarrow \quad x = 7$

So the graph dives down beside the line $x = 7$, and the domain is $x > 7$. A vertical asymptote is a vertical line, so it must have the form $x = \text{constant}$.

Switch to exponential form. The statement says the exponent taking $9$ to $x$ is $-\tfrac{3}{2}$.

$x = 9^{-3/2} = \frac{1}{9^{3/2}} = \frac{1}{\left(9^{1/2}\right)^3} = \frac{1}{3^3} = \frac{1}{27}$

A negative logarithm signals an argument between $0$ and $1$, never a negative argument: the output of a logarithm may be negative, but its input may not.

An $x$-intercept has height $0$, so set the logarithm equal to zero and switch forms.

$\log_5(2x - 3) = 0 \quad \Longrightarrow \quad 2x - 3 = 5^0 = 1$

Solve the linear equation.

$2x = 4 \quad \Longrightarrow \quad x = 2$

The intercept is $(2,\ 0)$. The value $x = \tfrac{3}{2}$ makes the argument zero, so it marks the vertical asymptote, not the intercept.

A logarithm needs a positive argument and a base that is positive and not equal to $1$.

The argument $\tfrac{1}{5}$ is positive, so $\log_3 \tfrac{1}{5}$ exists. Its value is negative, since $\tfrac{1}{5}$ lies between $0$ and $1$, but a negative value is perfectly legal.

$0 < \tfrac{1}{5} < 1 \quad \Longrightarrow \quad \log_3 \tfrac{1}{5} < 0$

The others all fail: no exponent gives $3^y = 0$ or $3^y = -9$, and a negative base such as $-3$ does not define an exponential function on the real numbers, so it cannot be inverted.

The hypothesis $\log_b 7 = c$ says exactly that $b^c = 7$.

Now use the exponent rule $b^{2c} = \left(b^c\right)^2$ and substitute.

$b^{2c} = \left(b^c\right)^2 = 7^2 = 49$

You never need the value of $b$ or $c$ separately: the pair is pinned together by the definition.