if log(x+1) 27=3 find the value of x

*(x+1) is the base

i know that the answer is 2 but i have no clue how to get that

log(x+1) 27=3

since,
logb N = x
b^x = N

this can be written as
(x + 1)^3 = 27
take cube root of each side
x + 1 = 3
x = 2

Thanks

To find the value of x in the equation log(x+1) 27 = 3, we can use the definition of logarithm. A logarithm is the exponent to which a base must be raised to obtain a certain number.

In this equation, the base is (x+1), and the number is 27. The exponent (= 3) represents the logarithm of the number.

Using the definition, we can rewrite the equation as:

27 = (x+1)^3

To solve for x, we need to isolate it on one side of the equation.

Taking the cube root of both sides to eliminate the exponent:

∛27 = ∛((x+1)^3)

Simplifying:

3 = x + 1

Subtracting 1 from both sides:

2 = x

Therefore, the value of x is 2.

To find the value of x in the equation log(x+1) 27 = 3, we can use the logarithmic property that states: log(base b) a = c is equivalent to b^c = a.

In this case, the base of the logarithm is (x+1), the result is 27, and the exponent is 3. We want to find the value of x.

So, we can rewrite the equation as: (x+1)^3 = 27.

Now, we can solve this equation by taking the cube root of both sides: ∛[(x+1)^3] = ∛27.

On the left side, the cube root and the cube will cancel out, leaving us with (x+1) on the left side.

On the right side, the cube root of 27 is 3, since 3^3 = 27.

Therefore, we have: x+1 = 3.

To solve for x, we simply subtract 1 from both sides of the equation: x = 3 - 1.

Thus, x = 2.

Therefore, the value of x in the equation log(x+1) 27 = 3 is 2.