DuckieDai Log Equation Solver Python Program

exponential functions · MIT License

Overview

Solve five common logarithmic equation patterns for x, including equations with a sum of two logs, and keep only answers that make every log defined.

Owner-tested on a TI calculator
Version 1.0.0
2455 bytes
Uses math

Download LOGEQ34.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
Every equation has the shape shown on its menu line; the program asks only for the numbers that fill it in.
A log written without a base works the same in any base, so options 2 to 4 do not ask for one.
Answers that are whole numbers or fractions with a bottom of 500 or less are shown exactly; others are rounded to six decimal places.

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Learn the math

How to solve logarithmic equations

The idea

Most log equations are solved the same way. Use the log rules to gather everything into one log on each side, then drop the logs and solve the equation that is left.

The catch is the domain. You can only take the log of a positive number, so an answer that makes any log's input zero or negative is not a solution, even when the algebra produces it. Such an answer is called extraneous, and this program leaves it out for you.

The formula

k*log_b(A) = log_b(A^k)

log_b(A) + log_b(B) = log_b(A*B)

log_b(A) - log_b(B) = log_b(A/B)

log_b(A) = c means A = b^c

b: the base: positive and not 1. Options 2 to 4 have no base to enter, because their answers are the same in every base
k: the number in front of a log; the first rule turns it into a power
h, j: the numbers added to x inside a log; for x - 2, enter -2
n, m: the positive numbers inside the logs that do not contain x
c: a plain number with no log around it

Using the program

Choose the menu line that matches your equation, then type the numbers that fill it in. Decimals and fractions such as 1/3 are accepted.
1 is k log_b(x+h) = log_b(n) + c. 2 is k log(x) = log(n). 3 is k log(x) - log(m) = log(n).
4 is log(x+h) = log(x) + log(n). 5 is log_b(x+h) + log_b(x+j) = c.
The answer appears as a solution set. Whole numbers and fractions are shown exactly, and options 1 to 4 add the decimal value underneath.
When no answer keeps every log defined, the program says NO VALID SOLUTION or NO SOLUTION instead.

Worked example

Solve log_2(x+2) + log_2(x-2) = 5. Choose 5 and enter base 2, h = 2, j = -2, and c = 5.

(x+2)(x-2) = 2^5 = 32, so x^2 = 36

The algebra gives x = 6 or x = -6. But x = -6 makes x + 2 = -4, and there is no log of a negative number, so the program shows SOLUTION SET: {6}.

Solve log(x+6) = log(x) + log(3). Choose 4 and enter h = 6 and n = 3. Combining the right side gives log(3x), so x + 6 = 3x, and the program shows SOLUTION SET: {3} and Decimal: 3.0.

Doing it by hand

Option 1 is the most involved pattern, because the plain number c has to be turned into a log before the logs can be combined. Every log is base 3 here.

Solve 2 log_3(x+1) = log_3(4) + 2.

2 = log_3(3^2) = log_3(9)

log_3((x+1)^2) = log_3(4) + log_3(9) = log_3(36)

(x+1)^2 = 36, so x + 1 = 6 or x + 1 = -6

x + 1 = -6 would put a negative number inside log_3, so only x + 1 = 6 counts, and x = 5. In the program, choose 1 and enter base 3, k = 2, h = 1, n = 4, and c = 2: it shows SOLUTION SET: {5} and Decimal: 5.0.

The program does the same thing in one line: it raises both sides to undo the logs, giving x + h = (n*b^c)^(1/k), and then checks that x + h is positive.

Tips and common mistakes

Rewrite your equation to match a menu line before you start, and keep the signs: log(x - 2) means h = -2.
Check every answer in the original equation. Any value that makes a log's input zero or negative must be thrown out.
log with no base usually means base 10 and ln means base e. For options 2 to 4 it does not matter which one your equation uses.
If the answer is not a simple fraction, the program rounds it to six decimal places.

Where this comes up

Solving log equations is a core skill in the exponential and logarithmic functions unit of algebra 2, precalculus, and college algebra, and it is usually where extraneous solutions are first taught.

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) Log Equation 3.4
# Guided equation patterns
from math import log,sqrt

def N(p):
 while 1:
  try:
   s=input(p).strip()
   if '/' in s:
    a,b=s.split('/')
    return float(a)/float(b)
   return float(s)
  except:print('Try number or a/b')

def F(x):
 if abs(x-round(x))<1e-8:return str(int(round(x)))
 for d in range(2,501):
  n=round(x*d)
  if abs(x-n/d)<1e-9:return str(n)+'/'+str(d)
 return str(round(x,6))

def done(x,valid=True):
 print('\n'*5)
 if not valid:print('NO VALID SOLUTION')
 else:
  print('SOLUTION SET:')
  print('{'+F(x)+'}')
  print('Decimal:',round(x,6))
 print('\n'*2)

print('DUCKIEDAI(TM)')
print('LOG EQUATION 3.4')
print('      __')
print('  ___(o )> quack!')
print("  \\ <_. )")
print("   `---'")
print('by DuckieDai')
print('hiddenwordscanner.com')
input('Press Enter ')
print('\n'*6)
while 1:
 print('1: klog_b(x+h)=log_b(n)+c')
 print('2: klog(x)=log(n)')
 print('3: klog(x)-log(m)=log(n)')
 print('4: log(x+h)=log(x)+log(n)')
 print('5: log_b(x+h)+log_b(x+j)=c')
 print('6: Quit')
 q=input('Choice? ')
 if q=='6':break
 if q not in ('1','2','3','4','5'):continue
 print('\n'*6)
 try:
  if q=='1':
   print('k log_b(x+h)=log_b(n)+c')
   b=N('Base b? ')
   k=N('k? ')
   h=N('h? ')
   n=N('n? ')
   c=N('c? ')
   if b<=0 or b==1 or k==0 or n<=0:
    print('Invalid base/input')
   else:
    v=(n*b**c)**(1/k)
    done(v-h,v>0)
  elif q=='2':
   print('k log(x)=log(n)')
   k=N('k? ')
   n=N('n? ')
   if k==0 or n<=0:print('Invalid input')
   else:done(n**(1/k))
  elif q=='3':
   print('klog(x)-log(m)=log(n)')
   k=N('k? ')
   m=N('m? ')
   n=N('n? ')
   if k==0 or m<=0 or n<=0:print('Invalid input')
   else:done((m*n)**(1/k))
  elif q=='4':
   print('log(x+h)=log(x)+log(n)')
   h=N('h? ')
   n=N('n? ')
   if n<=0:print('Invalid n')
   elif n==1:
    if h==0:print('ALL x > 0')
    else:print('NO SOLUTION')
   else:
    x=h/(n-1)
    done(x,x>0 and x+h>0)
  else:
   print('log_b(x+h)+log_b(x+j)=c')
   b=N('Base b? ')
   h=N('h? ')
   j=N('j? ')
   c=N('c? ')
   if b<=0 or b==1:print('Invalid base')
   else:
    disc=(h-j)**2+4*b**c
    roots=[(-(h+j)+sqrt(disc))/2, (-(h+j)-sqrt(disc))/2]
    good=[x for x in roots if x+h>0 and x+j>0]
    print('\n'*5)
    if good:
     print('SOLUTION SET:')
     print('{'+','.join(F(x) for x in good)+'}')
    else:print('NO SOLUTION')

 except (ValueError,OverflowError,ZeroDivisionError):
  print('Cannot solve input')
 input('Press Enter ')
 print('\n'*6)

# DuckieDai(TM) Log Equation 3.4
# Guided equation patterns
from math import log,sqrt
def N(p):
while 1:
try:
s=input(p).strip()
if '/' in s:
a,b=s.split('/')
return float(a)/float(b)
return float(s)
except:print('Try number or a/b')
def F(x):
if abs(x-round(x))<1e-8:return str(int(round(x)))
for d in range(2,501):
n=round(x*d)
if abs(x-n/d)<1e-9:return str(n)+'/'+str(d)
return str(round(x,6))
def done(x,valid=True):
print('\n'*5)
if not valid:print('NO VALID SOLUTION')
else:
print('SOLUTION SET:')
print('{'+F(x)+'}')
print('Decimal:',round(x,6))
print('\n'*2)
print('DUCKIEDAI(TM)')
print('LOG EQUATION 3.4')
print(' __')
print(' ___(o )> quack!')
print(" \\ <_. )")
print(" `---'")
print('by DuckieDai')
print('hiddenwordscanner.com')
input('Press Enter ')
print('\n'*6)
while 1:
print('1: klog_b(x+h)=log_b(n)+c')
print('2: klog(x)=log(n)')
print('3: klog(x)-log(m)=log(n)')
print('4: log(x+h)=log(x)+log(n)')
print('5: log_b(x+h)+log_b(x+j)=c')
print('6: Quit')
q=input('Choice? ')
if q=='6':break
if q not in ('1','2','3','4','5'):continue
print('\n'*6)
try:
if q=='1':
print('k log_b(x+h)=log_b(n)+c')
b=N('Base b? ')
k=N('k? ')
h=N('h? ')
n=N('n? ')
c=N('c? ')
if b<=0 or b==1 or k==0 or n<=0:
print('Invalid base/input')
else:
v=(n*b**c)**(1/k)
done(v-h,v>0)
elif q=='2':
print('k log(x)=log(n)')
k=N('k? ')
n=N('n? ')
if k==0 or n<=0:print('Invalid input')
else:done(n**(1/k))
elif q=='3':
print('klog(x)-log(m)=log(n)')
k=N('k? ')
m=N('m? ')
n=N('n? ')
if k==0 or m<=0 or n<=0:print('Invalid input')
else:done((m*n)**(1/k))
elif q=='4':
print('log(x+h)=log(x)+log(n)')
h=N('h? ')
n=N('n? ')
if n<=0:print('Invalid n')
elif n==1:
if h==0:print('ALL x > 0')
else:print('NO SOLUTION')
else:
x=h/(n-1)
done(x,x>0 and x+h>0)
else:
print('log_b(x+h)+log_b(x+j)=c')
b=N('Base b? ')
h=N('h? ')
j=N('j? ')
c=N('c? ')
if b<=0 or b==1:print('Invalid base')
else:
disc=(h-j)**2+4*b**c
roots=[(-(h+j)+sqrt(disc))/2, (-(h+j)-sqrt(disc))/2]
good=[x for x in roots if x+h>0 and x+j>0]
print('\n'*5)
if good:
print('SOLUTION SET:')
print('{'+','.join(F(x) for x in good)+'}')
else:print('NO SOLUTION')
except (ValueError,OverflowError,ZeroDivisionError):
print('Cannot solve input')
input('Press Enter ')
print('\n'*6)

Filename: LOGEQ34.py
Size: 2455 bytes
SHA-256: 2f8ea00b7e4aab3303c283ec003914446cc2f5abde8fba5aa1f2abcdb229c85d

Usage

Inputs

One menu choice from 1 to 6
Option 1: base b, k, h, n, and c
Option 2: k and n; option 3: k, m, and n
Option 4: h and n
Option 5: base b, h, j, and c

Outputs

The solution set, with exact whole-number or fraction answers where possible
The decimal value of the answer for options 1 to 4
NO VALID SOLUTION or NO SOLUTION when there is no answer, or a note that every positive x is a solution

Units: Pure numbers; no physical units.

Assumptions and limitations

Assumptions

Every equation has the shape shown on its menu line; the program asks only for the numbers that fill it in.
A log written without a base works the same in any base, so options 2 to 4 do not ask for one.
Answers that are whole numbers or fractions with a bottom of 500 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.
k cannot be 0, and the numbers inside the logs on the right side (n, and m in option 3) must be greater than 0.
Numbers may be typed as decimals or as a fraction such as 1/3; anything else asks again.

Known failures

An answer that breaks the domain of a log, such as a negative x + h, is reported as NO VALID SOLUTION or left out of the solution set.
Numbers too large for the calculator show Cannot solve input instead of an answer.
A menu choice other than 1 to 6 shows the menu again.

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: LOGEQ34 — 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 LOGEQ34 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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