Please help me because I do not any idea in Logarithm topic.Solve for the value of X: 6log(x^2+1)=5

6log(x^2 + 1) = 5

log(x^2 + 1 = 5/6
x^2+ 1 = 10^(5/6) , by definition of logs
x^2 = 10^5/6) - 1
x = ±√(10^(5/6 - 1)
= appr ± 2.41

check:
6 log(2.41^1 + 1)
= 6log 6.8081
= 6(.83302...) = 4.9981.. , not bad

Yes

To solve for the value of X in the equation 6log(x^2+1) = 5, we can follow these steps:

Step 1: Divide both sides of the equation by 6 to isolate the logarithm:
log(x^2+1) = 5/6

Step 2: Convert the logarithmic equation into an exponential equation by rewriting it in the form:
x^2 + 1 = 10^(5/6)

Step 3: Simplify the right-hand side of the equation:
x^2 + 1 = √(10^5)

Step 4: Square root both sides of the equation to remove the exponent:
√(x^2 + 1) = √(10^5)

Step 5: Solve for x by subtracting 1 from both sides of the equation:
x^2 = √(10^5) - 1

Step 6: Simplify the right-hand side of the equation using a calculator:
x^2 = 316.22776 - 1

Step 7: Subtract 1 from the square root result:
x^2 = 315.22776

Step 8: Take the square root of both sides to solve for x:
x = ±√(315.22776)

Therefore, the value of x can be either the positive or negative square root of 315.22776, which is approximately ±17.7449.