Baktik
Answered

Given that

[tex] \log_8(x+2) + \log_8 y = r [/tex]

and

[tex] \log_2(x-2) - \log_2 y = 2r + 1, [/tex]

show that

[tex] x^2 = 4 + 2 \times 32^r. [/tex]



Answer :

Let's solve the given system of logarithmic equations step-by-step.

Given equations:
1. [tex]\( \log_8(x + 2) + \log_8(y) = r \)[/tex]
2. [tex]\( \log_2(x - 2) - \log_2(y) = 2r + 1 \)[/tex]

Step 1: Simplifying the first equation using change of base formula

First, we use the change of base formula for logarithms:
[tex]\[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \][/tex]

For the first equation, simplify each logarithm using base 2:
[tex]\[ \log_8(x + 2) = \frac{\log_2(x + 2)}{\log_2(8)} \][/tex]
[tex]\[ \log_8(y) = \frac{\log_2(y)}{\log_2(8)} \][/tex]

Since [tex]\( \log_2(8) = 3 \)[/tex],
[tex]\[ \log_8(x + 2) = \frac{\log_2(x + 2)}{3} \][/tex]
[tex]\[ \log_8(y) = \frac{\log_2(y)}{3} \][/tex]

Thus, the first equation becomes:
[tex]\[ \frac{\log_2(x + 2)}{3} + \frac{\log_2(y)}{3} = r \][/tex]
Multiply both sides by 3 to simplify:
[tex]\[ \log_2(x + 2) + \log_2(y) = 3r \][/tex]

Step 2: Simplifying the second equation

The second equation is already in base 2:
[tex]\[ \log_2(x - 2) - \log_2(y) = 2r + 1 \][/tex]

Step 3: Solve the simplified equation for [tex]\( \log_2(y) \)[/tex]

From the first simplified equation:
[tex]\[ \log_2(x + 2) + \log_2(y) = 3r \][/tex]
Solve for [tex]\( \log_2(y) \)[/tex]:
[tex]\[ \log_2(y) = 3r - \log_2(x + 2) \][/tex]

Step 4: Substitute [tex]\( \log_2(y) \)[/tex] into the second equation

Substitute [tex]\( \log_2(y) \)[/tex] in the second equation:
[tex]\[ \log_2(x - 2) - (3r - \log_2(x + 2)) = 2r + 1 \][/tex]
Simplify the equation:
[tex]\[ \log_2(x - 2) - 3r + \log_2(x + 2) = 2r + 1 \][/tex]
Combine like terms:
[tex]\[ \log_2(x - 2) + \log_2(x + 2) - 3r = 2r + 1 \][/tex]

Move [tex]\( -3r \)[/tex] to the right:
[tex]\[ \log_2(x - 2) + \log_2(x + 2) = 5r + 1 \][/tex]

Step 5: Combine the logarithms using the product rule

Use the product rule for logarithms: [tex]\( \log_2(a) + \log_2(b) = \log_2(ab) \)[/tex]
[tex]\[ \log_2((x - 2)(x + 2)) = 5r + 1 \][/tex]

Simplify the argument of the logarithm:
[tex]\[ (x - 2)(x + 2) = x^2 - 4 \][/tex]
Thus, the equation becomes:
[tex]\[ \log_2(x^2 - 4) = 5r + 1 \][/tex]

Step 6: Solve for [tex]\( x^2 \)[/tex]

Rewrite the logarithmic equation in its exponential form:
[tex]\[ x^2 - 4 = 2^{(5r + 1)} \][/tex]

Therefore, adding 4 to both sides:
[tex]\[ x^2 = 4 + 2^{(5r + 1)} \][/tex]

Since [tex]\( 2^{(5r + 1)} = 2 \times 2^{5r} \)[/tex] and knowing that [tex]\( 2^5 = 32 \)[/tex]:
[tex]\[ x^2 = 4 + 2 \times 32^r \][/tex]

Thus, we have shown that:
[tex]\[ x^2 = 4 + 2 \times 32^r \][/tex]