DuckieDai Exponential Equation Solver Python Program

exponential functions · MIT License

Overview

Solve exponential equations of the form b^(a*x+c) = R step by step: match the exponents with a logarithm, solve for x, and check the answer.

Owner-tested on a TI calculator
Version 1.0.0
1340 bytes
Uses math

Download EXPEQ34.py Read the source How to install

Before you run it

TI-84 Plus CE Python · No calculator OS or Python App version has been verified.
TI Connect CE .py transfer to the Python App; the owner's transfer details were not recorded
The equation has the form b^(a*x+c) = R, with one base and one exponent that is linear in x.
Answers that are whole numbers or fractions with a bottom of 64 or less are shown exactly; others are rounded to six decimal places.
The base b must be greater than 0 and cannot be 1, and the right side R must be greater than 0.

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Learn the math

How to solve exponential equations

The idea

An exponential equation has x in the exponent, as in 2^(3x - 1) = 32. To solve it, bring the exponent down: take a logarithm with the same base on both sides, and the equation becomes an ordinary linear one.

When the right side is an exact power of the base, such as 32 = 2^5, the logarithm is a whole number and you can do it in your head. When it is not, the logarithm gives a decimal answer instead.

The formula

b^(a*x+c) = R means a*x + c = log_b(R)

log_b(R) = ln(R)/ln(b)

x = (log_b(R) - c)/a

b: the base: positive and not 1
a: the number multiplying x in the exponent
c: the number added in the exponent; for 3x - 1, c is -1
R: the right side, which must be positive

Using the program

Choose 1 to solve, then enter b, a, c, and R. Decimals and fractions such as 1/8 are accepted.
The first screen, MATCH EXPONENTS, shows a*x+c and a*x, the steps you would write by hand.
The SOLUTION screen shows x exactly when it is a whole number or a simple fraction, then the decimal, then a check of b^(a*x+c) using that x.
When a is 0 there is no x in the exponent, so the program says whether every x works or none does.

Worked example

Solve 2^(3x - 1) = 32. Choose 1 and enter b = 2, a = 3, c = -1, and R = 32.

3x - 1 = log_2(32) = 5, so 3x = 6

The program shows a*x+c = 5 and a*x = 6, then x = 2 with Decimal: 2.0 and Check: 32.0.

Solve 5^(2x) = 7, where 7 is not a power of 5. Enter b = 5, a = 2, c = 0, and R = 7. The program shows a*x+c = 1.209062, then x = 0.604531 and Check: 7.0.

Tips and common mistakes

Write the equation in the form b^(a*x+c) = R first. For 3^(x+2) = 1/9, b = 3, a = 1, c = 2, and R = 1/9, which gives x = -4.
A positive base raised to any power is positive, so a right side of 0 or a negative number has no solution.
Use the Check line: it should match R, apart from rounding.

Where this comes up

Exponential equations are solved in the exponential and logarithmic functions unit of algebra 2, precalculus, and college algebra, and the same steps solve growth, decay, and compound interest problems for time.

Go further: Logarithm Rules and How to Solve Exponential Equations

Exact reviewed bytes

Source code

Download .py

View source below, copy it when JavaScript is available, or download the exact .py bytes.

# DuckieDai(TM) Exp Eq 3.4
# Solve b^(a*x+c)=R
from math import log

def number(p):
 while 1:
  try:
   s=input(p).strip()
   if '/' in s:
    n,d=s.split('/')
    return float(n)/float(d)
   return float(s)
  except:
   print('Try 3, -2, 1/8')

def fmt(v):
 if abs(v-round(v))<1e-8:
  return str(int(round(v)))
 for d in range(2,65):
  n=round(v*d)
  if abs(v-n/d)<1e-8:
   return str(n)+'/'+str(d)
 return str(round(v,6))

def pause():
 input('Enter >')
 print('\n'*7)

print('DUCKIEDAI(TM)')
print('EXP EQUATION 3.4')
print('      __')
print('  ___(o )> quack!')
print('  \\ <_. )')
print("   `---'")
print('by DuckieDai')
print('hiddenwordscanner.com')
pause()
while 1:
 print('b^(a*x+c)=R')
 print('1 Solve  2 Quit')
 if input('Choice? ')!='1':break
 print('\n'*7)
 b=number('Base b? ')
 a=number('x coeff a? ')
 c=number('Constant c? ')
 r=number('Right side R? ')
 if b<=0 or b==1 or r<=0:
  print('Base >0, not 1')
  print('Right side >0')
 elif a==0:
  print('No x in exponent')
  print('All x work' if abs(b**c-r)<1e-8 else 'No solution')
 else:
  e=log(r)/log(b)
  x=(e-c)/a
  print('\n'*7)
  print('MATCH EXPONENTS')
  print('a*x+c =',fmt(e))
  print('a*x =',fmt(e-c))
  pause()
  print('SOLUTION')
  print('x =',fmt(x))
  print('Decimal:',round(x,6))
  print('Check:',round(b**(a*x+c),6))
 pause()
print('DuckieDai(TM) done!')

# DuckieDai(TM) Exp Eq 3.4
# Solve b^(a*x+c)=R
from math import log
def number(p):
while 1:
try:
s=input(p).strip()
if '/' in s:
n,d=s.split('/')
return float(n)/float(d)
return float(s)
except:
print('Try 3, -2, 1/8')
def fmt(v):
if abs(v-round(v))<1e-8:
return str(int(round(v)))
for d in range(2,65):
n=round(v*d)
if abs(v-n/d)<1e-8:
return str(n)+'/'+str(d)
return str(round(v,6))
def pause():
input('Enter >')
print('\n'*7)
print('DUCKIEDAI(TM)')
print('EXP EQUATION 3.4')
print(' __')
print(' ___(o )> quack!')
print(' \\ <_. )')
print(" `---'")
print('by DuckieDai')
print('hiddenwordscanner.com')
pause()
while 1:
print('b^(a*x+c)=R')
print('1 Solve 2 Quit')
if input('Choice? ')!='1':break
print('\n'*7)
b=number('Base b? ')
a=number('x coeff a? ')
c=number('Constant c? ')
r=number('Right side R? ')
if b<=0 or b==1 or r<=0:
print('Base >0, not 1')
print('Right side >0')
elif a==0:
print('No x in exponent')
print('All x work' if abs(b**c-r)<1e-8 else 'No solution')
else:
e=log(r)/log(b)
x=(e-c)/a
print('\n'*7)
print('MATCH EXPONENTS')
print('a*x+c =',fmt(e))
print('a*x =',fmt(e-c))
pause()
print('SOLUTION')
print('x =',fmt(x))
print('Decimal:',round(x,6))
print('Check:',round(b**(a*x+c),6))
pause()
print('DuckieDai(TM) done!')

Filename: EXPEQ34.py
Size: 1340 bytes
SHA-256: d73d13ec85d46e836a038ec7afaebe36aa3273415b24eeb111845a8253aa971a

Usage

Inputs

Menu choice: 1 to solve or 2 to quit
Base b
The coefficient a of x and the constant c in the exponent
The right side R

Outputs

The matched exponent a*x+c and the value of a*x
The solution x, exactly when possible, and as a decimal
A check of b^(a*x+c) using the solution

Units: Pure numbers; no physical units.

Assumptions and limitations

Assumptions

The equation has the form b^(a*x+c) = R, with one base and one exponent that is linear in x.
Answers that are whole numbers or fractions with a bottom of 64 or less are shown exactly; others are rounded to six decimal places.

Constraints

The base b must be greater than 0 and cannot be 1, and the right side R must be greater than 0.
Numbers may be typed as decimals or as a fraction such as 1/8; anything else asks again.

Known failures

When a is 0 there is no x in the exponent; the program reports whether every x works or none does.
A right side or base so large or small that the logarithm is infinite stops the program with an error.

Compatibility and review

Review status: Owner-tested on a TI calculator
Tested on: Owner reports personally running this program on a TI-84 Plus CE Python calculator; the test date, OS version, and Python App version were not recorded
What happened: Owner reported pass on the calculator. The file is published exactly as provided; this exact file has not yet been rerun on a calculator.
Dependencies: math
Suggested calculator name: EXPEQ34 — you can give it another valid, unique name when you transfer it
Desktop test cases: Not available for this interactive-only source

This exact source auto-launches an interactive calculator session, so compatibility evidence comes from the recorded physical-device review rather than an importable desktop fixture.

Transfer and launch

Download the .py file, send it to your calculator with TI Connect CE as EXPEQ34 or another name, then open the Python App and try the worked example on this page first. This release has no verified calculator OS or Python App version; treat transfer and execution as unverified until independently tested.

Full installation and troubleshooting guide. Check your teacher's or exam's rules before using any program.

Version history

1.0.0 — current version, published 2026-10-12. Download this reviewed version

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