Question

What is the value of x in the equation [log3(x+1)+log3(x2-x+1)]=2 ?
O 2​

Answer 1

Answer:

2

Step-by-step explanation:

we have log (x+1) ro base 3 + log (x² -x +1) to base 3 = 2

we have to find x

using property log(mn) = log m + log n

we have

log (x+1) ro base 3 + log (x² -x +1) to base 3

=log [(x+1) (x² - x +1)] to base 3 = 2

3² = (x + 1)(x²-x +1)

9 = x³ -x² + x + x² - x + 1

9 = x³ + 1

x³ = 8

x = 2

Answer 2

The value of x in the given equation is 2

Given,

[log3(x+1)+log3(x2-x+1)]=2

To Find,

The value of x

Solution,

We have been given the following equation:

${log}_{3} (x+1) + {log}_{3} (x^2 - x +1) = 2$

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

${log}_{3} ((x+1)(x^2 - x +1)) = 2\\\\\implies (x+1)(x^2 - x +1) = 3^2\\\\\implies x^3-x^2+x+x^2-x+1 = 9\\\\\implies x^3 + 1 = 9\\\\\implies x^3 = 8\\\\\implies x = 2$

Therefore, the value of x = 2

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