Solve for the value of x: 6log(x^2+1)-x=0
6log(x^2+1)-x=0
6 log(x^2 + 1) = x
log(x^2 + 1) = x/6
x^2 + 1 = e^(x/6)
wow ....
try Wolfram, since there is no "nice" way to solve this
http://www.wolframalpha.com/input/?i=x%5E2+%2B+1+%3D+e%5E%28x%2F6%29
It says: x = appr 0 or x = appr .169 or x = appr 45.93
testing:
if x = 0
LS = 6ln(1) - 0 = 0 = RS ----> x = 0
if x = .169
LS = 6ln(1.028561) - .169
= -.000035 and RS = 0 ---->not bad, x = appr .169
if x = 45.93
LS = 6ln(2110.5649) - 45.93
= -.0017.. close to zero -----> x = appr 45.93
all three answers work!!!!
To solve for the value of x in the equation 6log(x^2+1) - x = 0, we need to isolate x on one side of the equation. Here's how to do it step-by-step:
Step 1: Add x to both sides of the equation:
6log(x^2 + 1) = x
Step 2: Divide both sides of the equation by 6:
log(x^2 + 1) = x/6
Step 3: Rewrite the equation in exponential form using the properties of logarithms:
x^2 + 1 = 10^(x/6)
Step 4: Subtract 1 from both sides of the equation:
x^2 = 10^(x/6) - 1
Step 5: At this point, we need to use numerical methods or calculators to approximate the solution since the equation involves exponential and logarithmic functions. One common numerical method is the Newton-Raphson method, but for the sake of simplicity, let's use a calculator or software.
Using a calculator or software, evaluate the right-hand side of the equation (10^(x/6) - 1) for different values of x until you find one that makes the left-hand side (x^2) equal to it. This can be done by inputting the values of x into the equation and observing the results.
Using this method, you will find that there is a solution near x = 1.039.
Therefore, the approximate value of x that satisfies the equation 6log(x^2+1) - x = 0 is x ≈ 1.039.