To four decimal places, what is the approximate solution of e^2x + 3 = 7?

1.386
0.6931
2.0000
2.4730

Question

To four decimal places, what is the approximate solution of e^2x + 3 = 7?

	         1.386
 		 0.6931
 		 2.0000
 		 2.4730

campbell_st · Answer

so is the equation 

1. $$e^{2x} + 3 = 7$$

or 

2. 

$$e^{2x + 3} = 7$$

anonymous · Answer

1

campbell_st · Answer

ok... so subtract 3 from both sides... 

$$e^{2x} = 4$$

next take the base e log of both sides


$$\ln(e^{2x}) = \ln(4)$$

apply the low law for powers

$$2x \times \ln(e) = \ln(4)$$

now you need to know that that the log of the base is 1

$$\log_{a}(a) = 1$$

in this question you have 

$$\log_{e} (e) = \ln(e) = 1$$

so you problem now is 

$$2x \times 1 = \ln(4)$$

you should be able to solve it from here

anonymous · Answer

Okay ill try it

campbell_st · Answer

and check you answer by substituting back in...

anonymous · Answer

I cant figure it out. Can you show me farther?

campbell_st · Answer

ok... so you're happy 

$$2x = \ln(4)$$

then divide both sides by 2

$$x = \frac{\ln(4)}{2}$$

just evaluate that...