How do you solve for t $A = B a^{c t} + D$ $A= Ba^(ct)+D$ ?

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Answer 1 · 2018-05-07T14:56:43

$t = \frac{A - D}{B c ln a}$ $t=(A-D)/(Bclna)$

$A = B a^{c t} + D$ $A=Ba^(ct)+D$

$B a^{c t} = A - D$ $Ba^(ct)=A-D$

$a^{c t} = \frac{A - D}{B}$ $a^(ct)=(A-D)/B$

$c t ln a = \frac{A - D}{B}$ $ctlna=(A-D)/B$

$t = \frac{A - D}{B c ln a}$ $t=(A-D)/(Bclna)$

How do you solve for t A= Ba^(ct)+DA=Bact+DA= Ba^(ct)+D?