Solve for t using logarithms with base a?

Question

Please could someone help me with these problem and explain to me in detail, I really don't understand.....

1.) 2a^t/3=5
2.) K= H - Ca^t
3.)A= Ba^ct+D

Answer 1

1.) 2a^t/3=5

a^t/3=(5/2) ; and take log(base a)
log(base a) a^t/3 = log(base a)(5/2)

t/3 = log(base a)(5/2)
t = 3log(base a)(5/2) ANS...


2.) K= H - Ca^t

Ca^t= H - K
a^t =(H-K)/C ; and take log(base a)
t = log(base a)((H-K)/C) ANS...


3.)A= Ba^ct+D

Ba^ct = A-D
a^ct = (A-D) /B ; and take log(base a)
ct = log(base a)((A-D) /B)
t = (1/C)log(base a)((A-D) /B) ANS .....

Answer 2

1) 2a^t / 3 = 5

Multiply both sides by 3.

2a^t = 15

Divide both sides by 2.

a^t = (15/2)

Here's where you convert this into logarithmic form. Remember that

b^c = x in logarithmic form is log[base b](x) = c.

log[base a](15/2) = t

2) K = H - Ca^t

Isolate the a^t is the goal, in order to convert to logarithmic form.

K - H = -Ca^t

Divide both sides by -C.

(K - H)/(-C) = a^t

We can clean this up using the "negative one" technique. Factoring a (-1) out of the subtraction of two terms flips the terms.

(-1)(H - K)/(-C) = a^t

(H - K)/C = a^t

Convert to logarithmic form.

log[base a] ( (H - K)/C ) = t

3) A = Ba^(ct) + D

Goal: to isolate the exponential.

A - D = Ba^(ct)

(A - D)/B = a^(ct)

log[base a] ( (A - D)/B ) = ct

Divide both sides by c,

(1/c) log[base a] ( (A - D)/B ) = t

Answer 3

If a^x=b
Then logb to base a= x

Hence 2 a^t/3 =5

a^t/3=5/3

Therefore

log(5/3) to base a =t/3

or t= 3. log(5/3)



Similarly C a^t= H-K

a^t= (H-K)/C

t= log {(H-K)/C}

Note that above log is to base a

Ba^ct=A-D
a^ct=(A-D)/B

ct = log{(A-D)/B}
t= [log{(A-D)/B}]/c

Note that above log is to base a

Answer 4

Question 1
As written question leads to:-
2a^(t)= 15
a^(t) = 7.5
t log a = log(7.5)
t = log (7.5)
Perhaps question is meant to read as:-
2a^(t/3) = 5
a^(t/3) = 2.5
log(a^(t/3) = log(2.5)
(t/3) log a = log(2.5)
t/3 = log(2.5)
t = 3.log 5

Question 2
K = H - Ca^(t)
Ca^(t) = H - K
a^(t) = (H - K) / C
t = log[ (H - K) / C ]

Question 3
A = Ba^(ct) + D
a^(ct) = (A - D)/B
ct = log [ (A - D) / B ]
t = (1/c).log [ (A - D) / B]