How do I solve this exponential equation 5(e^x-2)*e^3x-9=150?

Question

campbell_st · Answer

well start by dividing both sides of the equation by 5

$$e^{x-2}\times e^{3x -9} = 30$$

now apply the index law for multiplying the same base... add the powers

which means 

$$e^{x -2 + 3x -9} = 30$$

simplify that

then take the base e log of both sides... 

this will allow you to solve for x. 

hope this helps

anonymous · Answer

I get .7522 as my answer.

campbell_st · Answer

nope... just doing a bit of mental arithmetic... the answer is closer to 5

anonymous · Answer

I got .9284 this time.

campbell_st · Answer

ok... taking the log of both sides 

$$\ln(e^{4x -11}) = \ln(30)$$

or 

$$4x -11 = \ln(30)$$

since 

$$\log_{a}(a^x) = x$$

that law had to be applied here...

anonymous · Answer

Would you add 11 to 30 then divide by 4?

campbell_st · Answer

add 11 to ln(30)

then divide the answer by 4

anonymous · Answer

3.600

anonymous · Answer

Is that closer to the answer?

campbell_st · Answer

thats a lot better

anonymous · Answer

Is that the correct answer?

campbell_st · Answer

well it is, given the way I worked it out...

campbell_st · Answer

you can check by substititing... is 

$$5 \times e^3.6 - 2) \times e^{3 \times 3.6 - 9} = 150...?$$

you'll get an approximation because of rounding... see if it works

campbell_st · Answer

oops should read 

$$5 \times e^{3.6 -2} \times e^{3 \times 3.6 - 9} = 150...?$$