SOLUTION: 1. Solve the following exponential equations a) 3^(2x-1) = 2^(x+2) b) 2^(x+1) = 5^(1-x) c) 2^(x+1)/ 5^x = 3 Can you please help me out? Thanks so much in advance:) Can you p

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: 1. Solve the following exponential equations a) 3^(2x-1) = 2^(x+2) b) 2^(x+1) = 5^(1-x) c) 2^(x+1)/ 5^x = 3 Can you please help me out? Thanks so much in advance:) Can you p      Log On


   

Question 744709: 1. Solve the following exponential equations
a) 3^(2x-1) = 2^(x+2)
b) 2^(x+1) = 5^(1-x)
c) 2^(x+1)/ 5^x = 3
Can you please help me out? Thanks so much in advance:)
Can you please show all the steps it would really help me understand:)

Answer by tommyt3rd(5050) About Me  (Show Source):
You can put this solution on YOUR website!
a)
3%5E%282x-1%29+=+2%5E%28x%2B2%29
Expanding exponents...
3%5E%282x%29%2A3%5E%28-1%29+=+2%5Ex%2A2%5E2
simplifying...
3%5E%282x%29%2F3+=+2%5Ex%2A4
%283%5E2%29%5Ex%2F3=4%2A2%5Ex
Next we isolate x...
9%5Ex%2F3=4%2A2%5Ex


9%5Ex%2F2%5Ex=4%2A3


Key step: we can combine the bases under the exponent since the exponent for both of them is the same...
%289%2F2%29%5Ex=12
log%28%289%2F2%29%5Ex%29=log12
x%2Alog%289%2F2%29=log12%29
x=log12%2Flog%289%2F2%29
which can be rewritten as...
log12%2F%28log9-log2%29

Less steps for the remaining parts...
b)
+2%5E%28x%2B1%29+=+5%5E%281-x%29
+2%5Ex%2A2+=+5%2A5%5E%28-x%29
2%5Ex%2F5%5Ex=5%2F2
%282%2F5%29%5Ex=5%2F2
%282%2F5%29%5Ex=5%2F2
%282%2F5%29%5Ex=%282%2F5%29%5E%28-1%29
after taking log of both sides and cancelling (i.e. if the bases are equal the the exponents are also equal to each other)


x=-1



Now with maximum efficiency we solve the last...
c)
2%5E%28x%2B1%29%2F+5%5Ex+=+3
%282%2A2%5Ex%29%2F5%5Ex=3
%282%2F5%29%5Ex=3%2F2

log%28%282%2F5%29%5Ex%29=log%283%2F2%29
x=%28log3-log2%29%2F%28log2-log5%29