Solve the equation
log[14](x+49) - log[196] x=1
Given
log14x+49 - log196x = 1
We will use a lemma that
logax=loga²=x²
Since 196=14², we write above as
log14x+49 - log14²x = 1
Then
log14²(x+49)² - log14²x = 1
Rewrite using laws of logarithm:
log14²(x+49)²/x = 1
Using the alternate form from the definition of logarithms,
(x+49)^2/x = 14²
Transpose and solve for x to get
(x-49)²=0, or
x=49
We will use a lemma that states:
logax = loga²x²
The proof of the lemma is left to you as an exercise, if it was not already covered in your course.
To solve the equation log[14](x+49) - log[196] x = 1, we can apply the logarithmic properties to simplify the equation.
The first step is to express the logarithms with the same base. We can rewrite log[14](x+49) and log[196] as log(x+49) / log(14) and log x / log(196), respectively.
Now we can rewrite the equation as:
log(x+49) / log(14) - log x / log(196) = 1
Next, we can use the property of logarithms that states log(a) - log(b) = log(a / b). Applying this property, we have:
log(x+49) / log(14) / (log x / log(196)) = 1
To simplify further, we can multiply both sides of the equation by log(14) / log(196):
log(x+49) / log x = log(14) / log(196)
Now we can use the property log(a) - log(b) = log(a / b) again to simplify further. Applying this property:
log((x+49) / x) = log(14/196)
Since the logarithm of a number is equal to the logarithm of another number if and only if the numbers themselves are equal, we can now remove the logarithms and solve the equation:
(x+49) / x = 14 / 196
Simplifying the right side of the equation gives:
(x+49) / x = 1 / 14
To eliminate the fractions, we can cross-multiply:
14(x+49) = x
Distributing 14 to the terms inside the parentheses gives:
14x + 686 = x
Moving the x term to the left side of the equation gives:
14x - x = -686
Combining like terms gives:
13x = -686
Dividing both sides of the equation by 13 gives the value of x:
x = -686 / 13
Therefore, the solution to the equation log[14](x+49) - log[196] x = 1 is x = -52.92 (rounded to two decimal places).