If bases cannot be made to be the same, then get in a = b form and take the natu
ID: 3123288 • Letter: I
Question
If bases cannot be made to be the same, then get in a = b form and take the natural log (ln) of both sides. Give EXACT answers (leave in terms of logarithms). a) Solve: 3e^t/3 - 2 = 8 b) Solve: e^2x - 3 = 5 c) If f(t) = 50e^-25t find t when f(t) = 10. If more than one term contains the variable in the exponent, try factoring. If factoring doesn't work, use u-substitution and the quadratic formula. a) Solve: e^2x - 5/e^-x - 2 = 0 Be able to solve log equations: Combine to a single log to get in log_b x = y form and then change it to exponential form: b^y = x. Give EXACT answers (answers may contain exponential terms). You MUST check your answers when solving log equations (check back into the original equation to make sure your potential solutions don't make you take log or natural log of a negative number). Solve: log (x - 1) = 2 - log (x + 1) Solve: log_2 (x + 1) - log_2(x - 1) = 3 Solve: ln (2x + 5) = 2Explanation / Answer
a.)
3e^t/3 - 2 =8
3e^t/3 = 10
tak log of both sides on the base e ,
ln3 + ln(e^t/3) = ln10
t/3 = ln10-ln3
t= 3(ln10-ln3)
t = 3( 2.3025 - 1.098)
t= 3.6135
b.) e^ 2x-3 =5
take log of both sides on the base e,
ln( e^2x-3) = ln5
2x-3 = ln5
2x = 3 + ln5
x =[ 3+ ln5 ]/2
x=2.3047
c.)
f(t) = 50* e ^ - .25t
because when f(t) = 10
10 = 50* e ^ -. 25t
take log of both sides on the base e ,
ln(10) = ln(50) -.25t
25t = ln(50)-ln(10)
.25t = 1.6094
t=6.437
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solve : log(x-1) = 2- log(x+1)
log(x-1) + log(x+1) = 2
log((x-1)(x+1))=2
log(x*x - 1)=2
x*x - 1 = e ^ 2
x*x = 1 + e^2
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solve : log2 (x+1) - log2 (x-1) = 3
log2 [ (x+1)/(x-1) ] =3
(x+1)/(x-1) = 2 ^ 3
(x+1)/(x-1) = 8
x + 1 = 8x -8
7x = 9
x = 9/7
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solve ln(2x+5) = 2
2x + 5 = e ^ 2
2x = (e ^ 2) -5
x = [(e ^ 2) -5]/2
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