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Solve the logarithmic equation $\log_{3}\left(x+3\right)+\log_{3}\left(x-3\right)=3$

Step-by-step Solution

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Solution

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Final answer to the problem

$x=6$
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Step-by-step Solution

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Learn how to solve factor by difference of squares problems step by step online. Solve the logarithmic equation log3(x+3)+log3(x+-3)=3. Express the numbers in the equation as logarithms of base 3. The sum of two logarithms of the same base is equal to the logarithm of the product of the arguments. Solve the product of difference of squares \left(x+3\right)\left(x-3\right). Use the following rule for logarithms: \log_b(b^k)=k.

Final answer to the problem

$x=6$

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Function Plot

Plotting: $\log_{3}\left(x+3\right)+\log_{3}\left(x-3\right)-3$

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(◻)
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◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

How to improve your answer:

Similar Problems Solved

$\log\left(x-1\right)+\log\left(x+1\right)=2\log\left(x+2\right)$

$\log x+\log\left(x-3\right)=1$

$2log\left(x\right)-log\left(x+6\right)=0$

$\log_x\left(64\right)=3$

$\log_2\left(x-2\right)+\log_2\left(x+1\right)=2$

$2\cdot log\left(x\right)-1\cdot log\left(x+6\right)=0$

$log\left(x+1\right)=log\left(x-1\right)+3$

Main Topic: Factor by Difference of Squares

The difference of two squares is a squared number subtracted from another squared number. Every difference of squares may be factored according to the identity a^2-b^2=(a+b)(a-b) in elementary algebra.

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