b. Log5x+log(x+1)=log5x

5 answers

To solve this equation, we can combine the logarithms using the properties of logarithms.

Using the property log(a) + log(b) = log(ab), we can rewrite the equation as:

log(5x(x + 1)) = log(5x)

Now, using the property log(a) = log(b) means a = b, we can remove the logarithms:

5x(x + 1) = 5x

Expanding the left side:

5x² + 5x = 5x

Subtracting 5x from both sides:

5x² = 0

Dividing both sides by 5:

x² = 0

Taking the square root of both sides:

x = 0

Therefore, the solution to the equation log5x + log(x + 1) = log5x is x = 0.
what a lot of work!
log5x+log(x+1)=log5x
just subtract log5x from both sides, and you have
log(x+1) = 0
x+1 = 1
x = 0
Apologies for the confusion in the previous response. You're absolutely right - we can simplify the equation by subtracting log5x from both sides.

Starting with log5x + log(x+1) = log5x, subtracting log5x from both sides gives us:

log(x+1) = 0

Since log5x - log5x = 0, we can simplify the equation to log(x+1) = 0.

Then, using the property log(b) = 0 resulting in b = 1, we have:

x + 1 = 1

Subtracting 1 from both sides:

x = 0

Therefore, the solution to the simplified equation log5x + log(x+1) = log5x is x = 0.
but in log (5x), x > 0 by definition of logs, so

there is no solution
Apologies for the incorrect response. You are correct. Upon closer examination, the equation log5x + log(x+1) = log5x does not have a valid solution.

By the rules of logarithms, we know that the argument of a logarithm must be greater than zero. If we consider the logarithm log5x in the equation, x must be greater than zero.

However, when x = 0, the equation becomes log(0+1) = 0, resulting in log(1) = 0. This is true, but it does not satisfy the original equation because x = 0 is not a valid input for log5x.

Therefore, there is no solution to the equation log5x + log(x+1) = log5x.