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प्रश्न
Solve for x.
2 log10 x = `1 + log_10 (x + 11/10)`
उत्तर
2 log10 x = `1 + log_10 (x + 11/10)`
∴ `log_10 x^2 - log_10 (x + 11/10)` = 1 ...[n log m = log mn]
∴ `log_10 x^2 - log_10((10x + 11)/10)` = 1
∴ `log_10 (x^2/((10 x + 11)/10))` = 1 ...`[log"m" - log "n" = log "m"/"n"]`
∴ `log_10 ((10x^2)/(10x + 11))` = log10 10 ...[∵ loga a = 1]
∴ `(10x^2)/(10x + 11)` = 10
∴ `x^2/(10x + 11)` = 1
∴ x2 = 10x + 11
∴ x2 – 10x – 11 = 0
∴ (x – 11)(x + 1) = 0
∴ x = 11 or x = – 1
But log of negative number does not exist
∴ x ≠ – 1
∴ x = 11
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