How do you solve $9^(n+10)+3=81$ ?

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Answer 1 · 2016-12-18T17:40:21

$n=-7/2$

$9^(n+10)+3=81$
$3^(2n+10)+3^1=3^4$
$2n+10+1=4$
$2n=4-11$
$2n=-7$
$n=-7/2$
substitute $2n=-7$
$-7+10+1=4$
$4=4$

Answer 2 · 2016-12-23T21:13:17

$n ~~ -8.017$

$9^(n+10) +3 = 81$

$9^(n+10) = 78" "$ the variable is in the index, so use logs

$log9^(n+10) = log78" "larr$ take logs of both sides

$(n+10)log9 = log78" "larr$ log law $loga^b = bloga$

$n+10= log78/log9$

$n = log78/log9 -10" "larr$ use a calculator at this point

$n = -8.017176$

How do you solve 9^(n+10)+3=819^(n+10)+3=81?