ðŸ‘‰ Try now NerdPal! Our new math app on iOS and Android

# Solve the logarithmic equation $\log \left(x\right)+\log \left(x-3\right)=1$

Go!
Symbolic mode
Text mode
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

##  Final answer to the problem

The equation has no solutions.

##  Step-by-step Solution 

How should I solve this problem?

• Choose an option
• Solve for x
• Find the derivative using the definition
• Solve by quadratic formula (general formula)
• Simplify
• Find the integral
• Find the derivative
• Factor
• Factor by completing the square
• Find the roots
Can't find a method? Tell us so we can add it.
1

Express the numbers in the equation as logarithms of base $10$

$\log \left(x\right)+\log \left(x-3\right)=\log \left(10^{1}\right)$

Learn how to solve differential equations problems step by step online.

$\log \left(x\right)+\log \left(x-3\right)=\log \left(10^{1}\right)$

Learn how to solve differential equations problems step by step online. Solve the logarithmic equation log(x)+log(x+-3)=1. Express the numbers in the equation as logarithms of base 10. Any expression to the power of 1 is equal to that same expression. The sum of two logarithms of the same base is equal to the logarithm of the product of the arguments. For two logarithms of the same base to be equal, their arguments must be equal. In other words, if \log(a)=\log(b) then a must equal b.

##  Final answer to the problem

The equation has no solutions.

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

###  Main Topic: Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives.