MHA5: A Proof and A Curve

Here at MHA5, I'll share two pretty good problems which involves logarithms. This time though, we won't get involved with logarithmic exponential equations, we're involved with something different.

Problem 1:
Let log_ax = α, log_bx = β, log_cx = γ, and log_dx = λ. Prove that log_abcdx = 1/(1/α + 1/β + 1/γ + 1/λ).

Okay, let's start off with expanding log_abcdx.
= log_abcdx
= logx / log(abcd)
= logx / (loga + logb + logc + logd)

Express loga in another form.
log_ax = α
a^α = x (Take the log of both sides)
log(a^α) = logx
αloga = logx
loga = logx/α

Expressing logb, logc and logd in another form is done using the same approach.
= logx / (logx/α + logx/β + logx/γ + logx/λ)
= logx / logx(1/α + 1/β + 1/γ + 1/λ)
= 1/(1/α + 1/β + 1/γ + 1/λ)

QED

Problem 2: 
Find a and k such that the graph y = a10^kx passes through the points (2, 6) and (5, 20). 

Let's substitute 2 and 5 in place of x and 6 and 20 in place of y in y = a10^kx.
6 = a10²^k and 20 = a10⁵^k

Express a as 6/(10²^k) and solve for k using the equation 20 = a10⁵^k .
20 = a10⁵^k
20 = 6/(10²^k) * 10⁵^k
20 = 6(10³^k)
20/6 = 10³^k (Take the log of both sides)
log(20/6) = log(10³^k)
log(20/6) = 3klog10
log(20/6) = 3k
(1/3)log(20/6) = k
0.1743 ≈ k

Solving for the value of a:
a = 6/(10²^k)
a = 6/(10²^{(1/3)log(20/6)})
a = 6/(10^{(2/3)log(20/6)})
a = 6/(10^log(20/6))^2/3

Since log(20/6) is equal to the number which should be the exponent of 10 to have 20/6, 10^{(2/3)log(20/6)} is just 20/6.

a = 6/(20/6)^2/3
a = 6/³√(100/9)
a ≈ 2.6888

I hope these problems entertained you. This is the end of my logarithm lessons, so see you next time.