Unlocking Solving Exponential Equations: A Comprehensive Guide

	Example Equation
With One Variable	$2^{x} = 32$
With the Same Variable on Both Sides	$2^{2 x} = 5 \cdot 2^{x}$
With the Same Base	$4^{3 x} = 4^{2 x + 3}$
With Unlike Bases	$3^{x + 4} = 8 1^{x}$
With a Rational Base	$(\frac{1}{2})^{x} = 8$

Example Equation

With One Variable

2^{x} = 32

With the Same Variable on Both Sides

2^{2 x} = 5 \cdot 2^{x}

With the Same Base

4^{3 x} = 4^{2 x + 3}

With Unlike Bases

3^{x + 4} = 8 1^{x}

With a Rational Base

(\frac{1}{2})^{x} = 8

Solution	Substitute	Simplify
$x = - 1$	$2 (3)^{- 1} = ? \frac{5}{3} (- 1) + \frac{7}{3}$	$\frac{2}{3} = \frac{2}{3} ✓$
$x \approx 0$	$2 (3)^{0} \approx ? \frac{5}{3} (0) + \frac{7}{3}$	$2 \approx \frac{7}{3} ⇕ 2 \approx 2.333333 \dots ✓$

Solution

Substitute

Simplify

x = - 1

2 (3)^{- 1} = ? \frac{5}{3} (- 1) + \frac{7}{3}

\frac{2}{3} = \frac{2}{3} ✓

x \approx 0

2 (3)^{0} \approx ? \frac{5}{3} (0) + \frac{7}{3}

2 \approx \frac{7}{3} ⇕ 2 \approx 2.333333 \dots ✓

a Since the exponential expressions on both sides of the equation already have the same base, the Property of Equality for Exponential Equations can be applied.

3^{3 x} = 3^{x + 1} \Leftrightarrow 3 x = x + 1

Next, the obtained equation can be solved for

x .

3 x = x + 1

Solve for

x

SubEqn

$LHS - x = RHS - x$

2 x = 1

DivEqn

$LHS / 2 = RHS / 2$

x = \frac{1}{2}

The solution will now be verified by substituting

\frac{1}{2}

for

x

in the given equation.

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

Substitute

$x = \frac{1}{2}$

3^{3 (\frac{1}{2})} = ? 3^{\frac{1}{2} + 1}

Evaluate

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

3^{\frac{3}{2}} = ? 3^{\frac{1}{2} + 1}

OneToFrac

Rewrite $1$ as $\frac{2}{2}$

3^{\frac{3}{2}} = ? 3^{\frac{1}{2} + \frac{2}{2}}

AddFrac

Add fractions

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

CalcPow

Calculate power

5.196152 \dots = 5.196152 \dots ✓

Since a true statement was obtained,

x = \frac{1}{2}

is a solution to the equation.

b The right-hand side of this equation must be written as a power of

5 .

2 5^{2 x + 1}

Rewrite

WritePow

Write as a power

(5^{2})^{2 x + 1}

PowPow

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

5^{2 (2 x + 1)}

Distr

Distribute $2$

5^{4 x + 2}

Now both sides of the equation are written as powers of

5 .

5^{x} = 2 5^{2 x + 1} \Leftrightarrow 5^{x} = 5^{4 x + 2}

Therefore, the Property of Equality for Exponential Equations will be used.

5^{x} = 5^{4 x + 2} \Leftrightarrow x = 4 x + 2

The value of

x

can be found by solving the equation.

x = 4 x + 2

Solve for

x

SubEqn

$LHS - 4 x = RHS - 4 x$

- 3 x = 2

DivEqn

$LHS / (- 3) = RHS / (- 3)$

x = \frac{2}{- 3}

MoveNegDenomToFrac

Put minus sign in front of fraction

x = - \frac{2}{3}

The solution can now be checked by substituting

- \frac{2}{3}

for

x

in the given equation.

5^{x} = 2 5^{2 x + 1}

Substitute

$x = - \frac{2}{3}$

5^{- \frac{2}{3}} = ? 2 5^{2 (- \frac{2}{3}) + 1}

Evaluate

MultPosNeg

$a (- b) = - a \cdot b$

5^{- \frac{2}{3}} = ? 2 5^{- 2 (\frac{2}{3}) + 1}

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

5^{- \frac{2}{3}} = ? 2 5^{- \frac{4}{3} + 1}

OneToFrac

Rewrite $1$ as $\frac{3}{3}$

5^{- \frac{2}{3}} = ? 2 5^{- \frac{4}{3} + \frac{3}{3}}

AddFrac

Add fractions

5^{- \frac{2}{3}} = ? 2 5^{- \frac{1}{3}}

CalcPow

Calculate power

0.341995 \dots = 0.341995 \dots ✓

A true statement was obtained. Therefore,

x = - \frac{2}{3}

is a solution to the equation.

c In this equation, the expressions on both sides will be written as powers of

2 .

To do so, some Properties of Exponents will be used. The left-hand side will be rewritten first.

2 (2)^{x - 3}

Rewrite

MultBasePow

$a \cdot a^{m} = a^{1 + m}$

2^{1 + x - 3}

SubTerm

Subtract term

2^{x - 2}

Next, the right-hand side of the equation will be rewritten as a power of

2 .

(\frac{1}{8})^{\frac{1}{3} x}

Rewrite

WritePow

Write as a power

(\frac{1}{2 ^{3}})^{\frac{1}{3} x}

FracToNegExponent

$\frac{1}{a ^{m}} = a^{- m}$

(2^{- 3})^{\frac{1}{3} x}

PowPow

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

2^{- 3 (\frac{1}{3} x)}

DenomMultFracToNumber

$3 \cdot \frac{a}{3} = a$

2^{- 1 x}

MultNegPos

$(- a) b = - a b$

2^{- x}

The expressions on both sides of the equation are now written as powers of the same base.

2 (2)^{x - 3} = (\frac{1}{8})^{\frac{1}{3} x} \Leftrightarrow 2^{x - 2} = 2^{- x}

Therefore, the Property of Equality for Exponential Equations can be used.

2^{x - 2} = 2^{- x} \Leftrightarrow x - 2 = - x

Finally, the obtained equation can be solved for

x .

x - 2 = - x

Solve for

x

AddEqn

$LHS + x = RHS + x$

2 x - 2 = 0

AddEqn

$LHS + 2 = RHS + 2$

2 x = 2

DivEqn

$LHS / 2 = RHS / 2$

x = 1

Now,

x = 1

will be checked by substitution.

2 (2)^{x - 3} = (\frac{1}{8})^{\frac{1}{3} x}

Substitute

$x = 1$

2 (2)^{1 - 3} = ? (\frac{1}{8})^{\frac{1}{3} (1)}

Evaluate

SubTerm

Subtract term

2 (2)^{- 2} = ? (\frac{1}{8})^{\frac{1}{3} (1)}

IdPropMult

Identity Property of Multiplication

2 (2)^{- 2} = ? (\frac{1}{8})^{\frac{1}{3}}

CalcPow

Calculate power

2 (\frac{1}{4}) = ? \frac{1}{2}

MoveLeftFacToNumOne

$a \cdot \frac{1}{b} = \frac{a}{b}$

\frac{2}{4} = ? \frac{1}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{1}{2} = \frac{1}{2} ✓

A true statement was obtained. Therefore,

x = 1

is a solution to the equation.

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Catch-Up and Review

Hint

Solution

Hint

Solution

Hint

Solution

Proof

If bx=by, Then x=y

If x=y, Then bx=by

Hint

Solution

Hint

Solution

Hint

Solution

If $b^{x} = b^{y},$ Then $x = y$

If $x = y,$ Then $b^{x} = b^{y}$