How do you solve $6^{- t} + 1 = 22$ $6^-t+1=22$ ?

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Answer 1 · 2016-12-28T19:08:54

$t = 1.700 (3 s . f .)$ $t= 1.700 (3s.f.)$

$6^{- t} + 1 = 22$ $6^(-t)+1=22$

subtract 1:

$6^{- t} = 21$ $6^(-t)=21$

convert to logarithmic form:

$a^{m} = n \to {log}_{a} (n) = m$ $a^m=n -> log_a(n)=m$

$6^{- t} = 21 \to {log}_{6} (21) = - t$ $6^(-t)=21 -> log_6(21)=-t$

solve using a calculator:

$log_6(21)= 1.69918...$

$-t = 1.69918...$

multiply by -1:

$t= 1.700 (3s.f.)$

How do you solve 6^-t+1=226−t+1=226^-t+1=22?