Certainly! Let's break down the problem step-by-step.
Given:
[tex]\[ x = \log(\sqrt{125}) \][/tex]
We need to find the value of [tex]\( x^2 + 1 \)[/tex].
### Step-by-Step Solution
1. Calculate [tex]\( \sqrt{125} \)[/tex]:
Recall that 125 can be factored as:
[tex]\[ 125 = 5^3 \][/tex]
Therefore,
[tex]\[ \sqrt{125} = \sqrt{5^3} = 5^{3/2} \][/tex]
Thus, the square root of 125 is approximately:
[tex]\[ \sqrt{125} \approx 11.180339887498949 \][/tex]
2. Calculate [tex]\( x = \log(\sqrt{125}) \)[/tex]:
We know that:
[tex]\[ x = \log(\sqrt{125}) \][/tex]
Substituting the value of [tex]\( \sqrt{125} \)[/tex] we get:
[tex]\[ x = \log(11.180339887498949) \][/tex]
The logarithm (in natural logarithm) of this value is approximately:
[tex]\[ x \approx 2.4141568686511508 \][/tex]
3. Calculate [tex]\( x^2 \)[/tex]:
Now, we square the value of [tex]\( x \)[/tex]:
[tex]\[ x^2 \approx (2.4141568686511508)^2 \approx 5.828153386455529 \][/tex]
4. Calculate [tex]\( x^2 + 1 \)[/tex]:
Finally, add 1 to [tex]\( x^2 \)[/tex]:
[tex]\[ x^2 + 1 \approx 5.828153386455529 + 1 = 6.828153386455529 \][/tex]
Therefore, the value of [tex]\( x^2 + 1 \)[/tex] is approximately:
[tex]\[ 6.828153386455529 \][/tex]
