Step-by-step explanation:
Given: p = [5/{(1/log₂x) + (1/log₃x) + (1/log₄x) + (1/log₅x)}]
⇛p = {5/(logₓ2 + logₓ3 +logₓ4 + logₓ5)
[since, 1/(log b^a) = log_(a) ᵇ]
⇛p = [5/{logₓ(2*3*4*5)}]
[since, logₓa + +logₓb = +logₓ(ab)]
⇛p = [5/{logₓ(6 * 20)}]
⇛p = {5/(logₓ120)}
⇛p = 5{1/(logₓ120)}
⇛p = 5 log₁₂₀x
[since, 1/log_(b) a = logₐb]
⇛p = log₁₂₀x⁵ →→→Eqn(1)
[since, m log a = log aᵐ]
And given: (120)ᵖ = 32
Now, from equation (1) substitute the value of p in this equation then
⇛(120)^(log₁₂₀x⁵) = 32
⇛x⁵ log₁₂₀120 = 32
[since, x log_(b) a = a log _(b) x]
⇛x⁵ * 1 = 32
[since, log_(a) a = 1]
⇛x⁵ = 32
⇛x⁵ = 2 * 2* 2* 2* 2
⇛x⁵ = 2⁵
Therefore, x = 2.
Answer: Hence, the value of x for the given problem is 2.
Please let me know if you have any other questions.
The value of x in [tex]p = \frac{5}{(1/log_2x) + (1/log_3x) + (1/log_4x) + (1/log_5x)}[/tex] is 2
The given parameters are:
[tex]p = \frac{5}{(1/log_2x) + (1/log_3x) + (1/log_4x) + (1/log_5x)}[/tex]
In logarithms, we have
1/(log b^a) = log_(a)ᵇ
So, the equation becomes
[tex]p = \frac{5}{log_x(2*3*4*5)}[/tex]
Also, we have the following logarithm rule
logₓa + +logₓb = +logₓ(ab)
So, the equation becomes
[tex]p = \frac{5}{log_x(6 * 20)}[/tex]
Evaluate the product
[tex]p = \frac{5}{log_x(120)}[/tex]
Apply the change of base rule
[tex]p = 5 \log_{120}x[/tex]
The equation becomes
[tex]p = \log_{120}x^5[/tex]
Recall that:
[tex](120)^p = 32[/tex]
Express as logarithm
[tex]p = \log_{120}32[/tex]
By comparing [tex]p = \log_{120}32[/tex] and [tex]p = \log_{120}x^5[/tex], we have:
[tex]x^5= 32[/tex]
Express 32 as an exponent
[tex]x^5= 2^5[/tex]
By comparison;
[tex]x = 2[/tex]
Hence, the value of x is 2
Read more about logarithmic equations at:
https://brainly.com/question/13473114
