Evaluate whether like terms can be combined across transcendental bases by fully simplifying: $7 \cdot 2^{x+1} - 5 \cdot 2^x + 3 \cdot 2^{x+2} - 4 \cdot 3^x + 2 \cdot 3^{x+1}$. Express your final answer in the form $A \cdot 2^x + B \cdot 3^x$, where $A$ and $B$ are integers.

Evaluate whether like terms can be combined across transcendental bases by fully simplifying: $7 \cdot 2^{x+1} - 5 \cdot 2^x + 3 \cdot 2^{x+2} - 4 \cdot 3^x + 2 \cdot 3^{x+1}$ . Express your final answer in the form $A \cdot 2^x + B \cdot 3^x$ , where $A$ and $B$ are integers.

Final answer

$21 \cdot 2^x + 2 \cdot 3^x$

Steps

Rewrite terms using exponent rules

To simplify the expression, we use exponent rules to isolate the common exponential factors $2^x$ and $3^x$ , and then combine like terms.

Recall the product rule for exponents: $a^{m+n} = a^m \cdot a^n$ .

Apply this rule to expand terms with shifted exponents:

For the terms involving base $2$ : $7 \cdot 2^{x+1} = 7 \cdot (2^x \cdot 2^1) = 7 \cdot 2 \cdot 2^x = 14 \cdot 2^x$ $-5 \cdot 2^x = -5 \cdot 2^x$ $3 \cdot 2^{x+2} = 3 \cdot (2^x \cdot 2^2) = 3 \cdot 4 \cdot 2^x = 12 \cdot 2^x$
For the terms involving base $3$ : $-4 \cdot 3^x = -4 \cdot 3^x$ $2 \cdot 3^{x+1} = 2 \cdot (3^x \cdot 3^1) = 2 \cdot 3 \cdot 3^x = 6 \cdot 3^x$

Group and combine like terms for each base

Substitute the rewritten terms back into the original expression: $(14 \cdot 2^x - 5 \cdot 2^x + 12 \cdot 2^x) + (-4 \cdot 3^x + 6 \cdot 3^x)$

Factor out $2^x$ and $3^x$ from their respective groups: $(14 - 5 + 12) \cdot 2^x + (-4 + 6) \cdot 3^x$

Calculate the integer coefficients: $14 - 5 + 12 = 21$ $-4 + 6 = 2$

This yields: $21 \cdot 2^x + 2 \cdot 3^x$

Check whether terms across different bases can be combined

Terms with different bases, such as $2^x$ and $3^x$ , are not like terms because their variable components grow at different exponential rates and are linearly independent functions. Therefore, they cannot be merged into a single term.

The fully simplified expression in the form $A \cdot 2^x + B \cdot 3^x$ is: $21 \cdot 2^x + 2 \cdot 3^x$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student incorrectly adds bases across different exponential bases, combining 21 · 2^x and 2 · 3^x into 23 · 5^x.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
= 7 $\cdot$ 2 $\cdot$ 2^x - 5 $\cdot$ 2^x + 3 $\cdot$ 4 $\cdot$ 2^x - 4 $\cdot$ 3^x + 2 $\cdot$ 3 $\cdot$ 3^x $\implies$ = 14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 12 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x
Step 2
= 21 $\cdot$ 2^x + 2 $\cdot$ 3^x
Step 3
= 23 $\cdot$ 5^x
Correct
21 $\cdot$ 2^x + 2 $\cdot$ 3^x
You attempted to add the coefficients ( $21 + 2 = 23$ ) and add the bases ( $2 + 3 = 5$ ). Terms with different bases, such as $2^x$ and $3^x$ , are not like terms and cannot be merged into a single power.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly used exponent rules to break down $2^{x+1}$ , $2^{x+2}$ , and $3^{x+1}$ .

You correctly simplified and collected like terms sharing the same base to obtain $21 \cdot 2^x + 2 \cdot 3^x$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Combining unlike exponential bases

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated terms with different exponential bases ( $2^x$ and $3^x$ ) as combinable like terms by simultaneously adding their coefficients and their bases.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check for like terms in exponential expressions: Two exponential terms $A \cdot a^x$ and $B \cdot b^x$ are like terms if and only if they share both the exact same base ( $a = b$ ) and exponent. If $a \neq b$ , they cannot be combined into a single base term.
Leave terms with different bases separate: $21 \cdot 2^x + 2 \cdot 3^x$ is already in simplest form.
Respect the requested form: When a problem specifies the target form $A \cdot 2^x + B \cdot 3^x$ , the final answer must retain both bases $2^x$ and $3^x$ .
Stop simplifying once the expression matches the required template $A \cdot 2^x + B \cdot 3^x$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Recognize that $21 \cdot 2^x$ and $2 \cdot 3^x$ involve linearly independent functions with different bases ( $2 \neq 3$ ).
Verify that the expression is already in the requested form $A \cdot 2^x + B \cdot 3^x$ , where $A = 21$ and $B = 2$ .

Final answer

21 $\cdot$ 2^x + 2 $\cdot$ 3^x

NEXT STEPS

Final Takeaways from Your Teacher

Review the definition of like terms to reinforce why $a^x$ and $b^x$ cannot be merged when $a \neq b$ .
Practice simplifying mixed exponential expressions and leaving them in partitioned-base form.

OverviewACEpath Teacher's Diagnosis

02 The student misapplies the exponent rule by adding the shift to the coefficient instead of multiplying by the base raised to the power, treating 2^(x+1) as 2^x + 1.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
7 $\cdot$ 2^x + 14 - 5 $\cdot$ 2^x + 3 $\cdot$ 2^x + 12 - 4 $\cdot$ 3^x + 2 $\cdot$ 3^x + 6
Correct
14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 12 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x
You expanded $2^{x+1}$ as $2^x + 2$ , $2^{x+2}$ as $2^x + 4$ , and $3^{x+1}$ as $3^x + 3$ . By the product rule of exponents, adding terms in an exponent corresponds to multiplication of powers: $b^{x+n} = b^x \cdot b^n$ . Therefore, $7 \cdot 2^{x+1} = 7 \cdot (2^1 \cdot 2^x) = 14 \cdot 2^x$ , $3 \cdot 2^{x+2} = 3 \cdot (2^2 \cdot 2^x) = 12 \cdot 2^x$ , and $2 \cdot 3^{x+1} = 2 \cdot (3^1 \cdot 3^x) = 6 \cdot 3^x$ .
Step 2
5 $\cdot$ 2^x - 2 $\cdot$ 3^x + 32
Correct
21 $\cdot$ 2^x + 2 $\cdot$ 3^x
Your arithmetic $(7 - 5 + 3 = 5$ , $-4 + 2 = -2$ , and $14 + 12 + 6 = 32$ ) correctly combined the terms from your previous line, but the result is affected by the incorrect expansion in Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You recognized that terms with different exponential bases like $2^x$ and $3^x$ cannot be merged together.

You accurately collected like terms and calculated integer coefficients based on your intermediate expression.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Additive Exponent Misconception

You split $b^{x+n}$ into $b^x + b^n$ instead of multiplying factors as $b^x \cdot b^n$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You applied a distributive/additive model to exponents by treating addition in the exponent as polynomial addition outside the base ( $b^{x+n} \to b^x + b^n$ ). Because exponentiation distributes over multiplication rather than addition, an addition in the exponent always translates into multiplying by the base raised to that power.

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent Product Law: For any positive base $b$ and real exponents, $b^{x+n} = b^x \cdot b^n$ , never $b^x + b^n$ .
Whenever you see a sum in an exponent, immediately factor it out as a product before distributing any outer coefficient.
Scaling Coefficients: $c \cdot b^{x+n} = c \cdot (b^n \cdot b^x) = (c \cdot b^n) \cdot b^x$ .
Calculate the constant multiplier $c \cdot b^n$ first, keeping $b^x$ as the sole variable factor.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite each shifted exponential term using $b^{x+n} = b^n \cdot b^x$ :
$7 \cdot 2^{x+1} = 7 \cdot (2^1 \cdot 2^x) = 14 \cdot 2^x$
$3 \cdot 2^{x+2} = 3 \cdot (2^2 \cdot 2^x) = 3 \cdot 4 \cdot 2^x = 12 \cdot 2^x$
$2 \cdot 3^{x+1} = 2 \cdot (3^1 \cdot 3^x) = 2 \cdot 3 \cdot 3^x = 6 \cdot 3^x$
Group like terms for each base:
$(14 - 5 + 12) \cdot 2^x + (-4 + 6) \cdot 3^x$
Simplify the integer coefficients: $14 - 5 + 12 = 21$ and $-4 + 6 = 2$ .

Final answer

21 $\cdot$ 2^x + 2 $\cdot$ 3^x

NEXT STEPS

Final Takeaways from Your Teacher

Review the product rule of exponents: verify that $2^{x+1} = 2 \cdot 2^x$ by testing a small integer value like $x = 2$ .
Practice converting expressions of the form $k \cdot b^{x+c}$ into $A \cdot b^x$ before attempting multi-term additions.
Complete the three targeted follow-up exercises below to build confidence in separating base multipliers.

OverviewACEpath Teacher's Diagnosis

03 The student evaluates 2^(x+2) as 2 · 2^x rather than 4 · 2^x, confusing 2^2 with 2.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 3 $\cdot$ 2 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x $\implies$ 14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 6 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x
Correct
14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 3 $\cdot$ 4 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x $\implies$ 14 $\cdot$ 2^x - 5 $\cdot$ 2^x + 12 $\cdot$ 2^x - 4 $\cdot$ 3^x + 6 $\cdot$ 3^x
You expanded $7 \cdot 2^{x+1} = 14 \cdot 2^x$ and $2 \cdot 3^{x+1} = 6 \cdot 3^x$ accurately. However, when splitting $2^{x+2}$ , the exponent rule gives $2^{x+2} = 2^x \cdot 2^2 = 4 \cdot 2^x$ , not $2 \cdot 2^x$ .
Step 2
15 $\cdot$ 2^x + 2 $\cdot$ 3^x
Correct
21 $\cdot$ 2^x + 2 $\cdot$ 3^x
Your arithmetic $(14 - 5 + 6 = 15)$ and $(-4 + 6 = 2)$ followed logically from your prior line, but the result for the $2^x$ term is affected by the earlier exponent slip.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized that terms with like exponential factors ( $2^x$ and $3^x$ ) should be grouped separately.

You accurately applied the product rule of exponents to $7 \cdot 2^{x+1}$ and $2 \cdot 3^{x+1}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Exponent Evaluation Slip

You wrote $2^{x+2}$ as $2 \cdot 2^x$ , forgetting to evaluate the power $2^2 = 4$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the constant shift in the exponent with a constant multiplier, or writing the base without squaring it when splitting $b^{x+n} = b^n \cdot b^x$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent Separation Rule: $b^{x+n} = b^x \cdot b^n$
Always compute the numerical constant $b^n$ before multiplying it by the outside coefficient.
Evaluate Powers First: $2^2 = 4$ , $2^3 = 8$ , $3^2 = 9$
Check that the separated numerical factor reflects the base raised to the exponent shift, not just the shift itself.

CORRECTED PATH

ACEpath Teacher's Demonstration

Rewrite $3 \cdot 2^{x+2}$ as $3 \cdot (2^2 \cdot 2^x) = 3 \cdot 4 \cdot 2^x = 12 \cdot 2^x$ .
Combine the coefficients for base $2$ : $(14 - 5 + 12) \cdot 2^x = 21 \cdot 2^x$ .
Combine the coefficients for base $3$ : $(-4 + 6) \cdot 3^x = 2 \cdot 3^x$ .

Final answer

21 $\cdot$ 2^x + 2 $\cdot$ 3^x

NEXT STEPS

Final Takeaways from Your Teacher

Practice separating exponential terms with shifts greater than $1$ , such as $a^{x+2}$ and $a^{x+3}$ .
Write out the step $b^n$ explicitly before combining coefficients.

Lock it in

Question

Can you combine $A \cdot 2^x + B \cdot 3^x$ into a single exponential term?

Answer

No. Terms with different exponential bases are not like terms, so their coefficients and bases cannot be added together.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student incorrectly adds bases across different exponential bases, combining 21 · 2^x and 2 · 3^x into 23 · 5^x.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student misapplies the exponent rule by adding the shift to the coefficient instead of multiplying by the base raised to the power, treating 2^(x+1) as 2^x + 1.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student evaluates 2^(x+2) as 2 · 2^x rather than 4 · 2^x, confusing 2^2 with 2.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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