Solve exponential equations and explain please?

1.) 3^(4x) = 27

3^(4x) = 3^3 .. 27 = 3^3

4x = 3 ... equate the exponents

x = 3/4

2.) e^(2x) = 4 ......... take ln of each side

2x = ln(4)

x = ln(4) / 2

3.) 3e^(3x) = 12 ... divide by 3

e^(3x) = 4 ...... take ln of each side

3x = ln(4)

x = ln(4) / 3

4.)10^(2x-3) +3 = 19 .. subtract 3

10^(2x-3) = 16 ... log (base 10) of each side

2x - 3 = log16

2x = log 16 + 3

x = (log(16) + 3) / 2

5.) 3e^x +7 = 9 ... subtract 7

3e^x = 2 ... divide 3

e^x = 2/3 ... take ln of both sides

x = ln(2/3)

x = ln 2 - ln 3

1: This one is easy, 27 = 3^3 so since the bases are the same, the exponents must be the same, x=3/4

2: Taking the natural logarithm of each side (because of base e) gives 2x=ln(4) so x=1/2ln(4) = ln(2) = 0.693 -- (know your log laws?)

3: Same as before but divide each side by 3. Gives 3x=ln(4) so x = 1/3ln(4) = ln(cubert(4)) = 0.462

4: First subtract 3 from both sides, then take log to the base 10 of each side gives 2x-3 = log10(16) so x = log10(4) +3/2

5: Same as before, e^x = 2/3, x = ln(2/3)

Hope this helps! Your calculator should have a button for log10 and ln on it (log10 usually called just log)

1. 3^4x =27

3^4x = 3^3

4x = 3

x = 3/4

2.e^2x = 4

2xlne =2ln2

x = ln2

3. 3e^3x = 12

e^3x = 4

3xlne = 2 ln2

x = (2/3)ln2

4.10^(2x-3) +3 =19

(2x-3)log 10 = log(2^4) = 4log2

2x - 3 = 4 log2

x = 2 log 2 + 3/2

5.3e^x +7 = 9

e^x = (9 -7)/3

x = ln(2/3)

1) 0.443

2) 0.693

3) 0.462

4) 2.102

5) -0.405