When solving an equation we should choose any method that makes things easier for us. In this exercise, we will show how to solve different kinds of exponential and logarithmic equations.

a. Solving $1 6^{x} = 2$

Note that this equation uses

16

as a base for the exponential expression in the left-hand side, while the one on the right-hand uses

2 .

Since

16 = 2^{4},

we can rewrite the equation so that both sides have the same base and we can equate the exponents. This because equal powers with same bases have same exponents. Let's give it a try.

1 6^{x} = 2

▼

Solve for

x

Rewrite

Rewrite $16$ as $2^{4}$

(2^{4})^{x} = 2

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

2^{4 x} = 2

$a = a^{1}$

2^{4 x} = 2^{1}

EquateExponents

Equate exponents

4 x = 1

DivEqn

$LHS / 4 = RHS / 4$

x = \frac{1}{4}

b. Solving $2^{x} = 4^{2 x + 1}$

This case is similar to the previous one, but here we have two exponential expressions on each side. Since

4 = 2^{2},

we can rewrite the right-hand side and simplify so that both exponential expressions have the same base. We can then equate the exponents.

2^{x} = 4^{2 x + 1}

▼

Solve for

x

Rewrite

Rewrite $4$ as $2^{2}$

2^{x} = (2^{2})^{2 x + 1}

PowPow

$(a^{m})^{n} = a^{m \cdot n}$

2^{x} = 2^{4 x + 2}

EquateExponents

Equate exponents

x = 4 x + 2

SubEqn

$LHS - 4 x = RHS - 4 x$

- 3 x = 2

DivEqn

$LHS / - 3 = RHS / - 3$

x = - \frac{2}{3}

c. Solving $2^{x} = 3^{x + 1}$

In this equation the bases involved do not have anything in common, as they are different prime numbers. We can use both sides as arguments of a logarithm and use the Properties of Logarithms to solve it.

2^{x} = 3^{x + 1}

▼

Solve for

x

$lo g (LHS) = lo g (RHS)$

lo g (2^{x}) = lo g (3^{x + 1})

$lo g (a^{m}) = m \cdot lo g (a)$

x lo g (2) = (x + 1) lo g (3)

Distr

Distribute $lo g (3)$

x lo g (2) = x lo g (3) + lo g (3)

SubEqn

$LHS - x lo g (2) = RHS - x lo g (2)$

0 = x lo g (3) - x lo g (2) + lo g (3)

FactorOut

Factor out $x$

0 = x (lo g (3) - lo g (2)) + lo g (3)

$lo g (m) - lo g (n) = lo g (\frac{m}{n})$

0 = x lo g (\frac{3}{2}) + lo g (3)

SubEqn

$LHS - lo g (3) = RHS - lo g (3)$

- lo g (3) = x lo g (\frac{3}{2})

DivEqn

$LHS / lo g (\frac{3}{2}) = RHS / lo g (\frac{3}{2})$

- \frac{lo g ( 3 )}{lo g ( \frac{3}{2} )} = x

RearrangeEqn

Rearrange equation

x = - \frac{lo g ( 3 )}{lo g ( \frac{3}{2} )}

d. Solving $lo g x = \frac{1}{2}$

In this case we can use the appropriate exponential function to undo the logarithmic function. Since the notation is just

lo g,

we know that this is a common logarithm and the base is

10 .

lo g x = \frac{1}{2}

▼

Solve for

x

$1 0^{LHS} = 1 0^{RHS}$

1 0^{l o g x} = 1 0^{\frac{1}{2}}

$1 0^{l o g (m)} = m$

x = 1 0^{\frac{1}{2}}

$a^{1 / 2} = a$

x = 10

e. Solving $ln x = 2$

Once more, we can use the appropriate exponential function to undo the logarithmic function. Since the notation is

ln

we know that this is a natural logarithm and the base is the natural base

e .

ln x = 2

▼

Solve for

x

$e^{LHS} = e^{RHS}$

e^{l n x} = e^{2}

$e^{l n (a)} = a$

x = e^{2}

f. Solving $lo g_{3} x = \frac{3}{2}$

Similarly as in the two previous cases we can use an exponential function to undo the logarithmic function. In this case the notation

lo g_{3}

lets us know that we should use

3

as a base.

lo g_{3} x = \frac{3}{2}

▼

Solve for

x

$3^{LHS} = 3^{RHS}$

3^{l o g_{3} x} = 3^{\frac{3}{2}}

$3^{l o g_{3} (m)} = m$

x = 3^{\frac{3}{2}}

$a^{m / n} = n a^{m}$

x = 3^{3}

CalcPow

Calculate power

x = 27

a. Solving 16x=2

b. Solving 2x=42x+1

c. Solving 2x=3x+1

d. Solving logx=21​

e. Solving lnx=2

f. Solving log3​x=23​

a. Solving $1 6^{x} = 2$

b. Solving $2^{x} = 4^{2 x + 1}$

c. Solving $2^{x} = 3^{x + 1}$

d. Solving $lo g x = \frac{1}{2}$

e. Solving $ln x = 2$

f. Solving $lo g_{3} x = \frac{3}{2}$

{{ 'ml-heading-exercise' | message }} {{ exercise.name }} {{ 'ml-label-page' | message }} {{ exercise.page }}

{{ 'ml-solutions-missing-heading' | message }}