Find the value of y, if log (y + 8) + log (y - 8) = 2log 3 + 2log 5

Analyze the question: We need to find y from a logarithmic equation. The key is to combine the logs on each side and then drop the logs.

Concept:

• log a + log b = log (ab)
• n log a = log aⁿ

Solution:

Left side: log (y + 8) + log (y − 8) = log[(y + 8)(y − 8)] = log (y² − 64)

Right side: 2 log 3 + 2 log 5 = log 9 + log 25 = log 225

So: log (y² − 64) = log 225

y² − 64 = 225

y² = 289

y = ±17

Checking validity: For the logs to be defined, we need y + 8 > 0 and y − 8 > 0, so y > 8. This means strictly only y = 17 works, since y = −17 would make both log(−9) and log(−25) undefined. However, the options are given in the "±" form, and y² = 289 gives ±17, so the intended answer is option C.

Let me also check the other options: y = ±13 gives y² − 64 = 105 (not 225), y = ±10 gives 36, y = ±5 gives −39. None of these work, so only ±17 fits.

Answer: C (y = ±17)

Tip: In log equations, always combine into a single log on each side, then equate the insides. Remember to check domain restrictions (arguments must be positive). In JAMB, if the options are in "±" form, go with the one matching y² from your working.