If 2^{x+y} = 32 and \log_3 x + \log_3 y = 1 + \log_3 2, what is the value of x^2 + y^2? — JAMB UTME

If $2^{x+y} = 32$ and $\log_3 x + \log_3 y = 1 + \log_3 2$ , what is the value of $x^2 + y^2$ ?

The value of

x^2 + y^2

13

Expressing

32

2^5

gives

x + y = 5

. Converting

1

\log_3 3

allows combining the right-hand side to

\log_3(3 \times 2) = \log_3 6

, so

\log_3(xy) = \log_3 6 \implies xy = 6

. Evaluating

x^2 + y^2 = (x+y)^2 - 2xy

yields

5^2 - 2(6) = 25 - 12 = 13

Simplify the exponential equation using index laws.

x + y = 5

Since

32 = 2^5

2^{x+y} = 2^5 \implies x + y = 5

Simplify the logarithmic equation using logarithm laws.

xy = 6

\log_3 x + \log_3 y = \log_3(xy)

and

1 + \log_3 2 = \log_3 3 + \log_3 2 = \log_3(3 \times 2) = \log_3 6

. Therefore,

xy = 6

Use the algebraic identity to find

x^2 + y^2

x^2 + y^2 = 13

Using

(x+y)^2 = x^2 + y^2 + 2xy

, we have

5^2 = x^2 + y^2 + 2(6) \implies 25 = x^2 + y^2 + 12 \implies x^2 + y^2 = 13

Solving simultaneous equations involving indices and logarithms using index laws and logarithmic identities.

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