Exponential Functions Problem?

Question

Can anyone help me solve for x on the following:

a.) 8^x=(1/32)^x-2

b.) 3^2x-12(3^x)+27=0

c.)(4x^3e^3x)-3x^4e^3x=0
**Without the exponents, the problem reads "4xe-3xe" its just hard to type that in an easy to read manner!

d.)e^2x=7

e.) 4^2x+3=5

If you can work out even just one problem, that would be great, but I would like to have examples of all problems at some point. Thanks a lot!!!!

For question e.), it should be typed as 4^(2x+3)=5

Thanks!

Answer 1

a.) do you mean

8^x=(1/32)^(x-2)? write everything in terms of 2

(2^3)^x = (1/2^5)^(x-2), right?

2^(3x) = (1/2^5)^(x-2), power to a power means you multiply

2^(3x) = (2^5)^(2-x), since flipping over means taking the negative exponent

2^(3x) = (2^(5(2-x)), again, power to a power means mult

so you see, 3x = 5(2-x) or
3x = 10-5x, gather
8x = 10, div by 8
x = 10/8 or
x = 5/4.

Please use parentheses to clarify the operations! I am not the only one having a hard time following you.

d.)e^(2x) =7, I THINK? Take ln of both sides

ln( e^(2x)) = ln(7),

(2x) ln( e) = ln(7), ln of a power, bring down the power
2x = ln(7), since ln(e) = 1
x = ln(7)/2.

Done.

Answer 2

The general rule is to logs of both sides, but this doesn't work too well if you have addition of plain-old constant terms. There, you need a little trial and error soft shoe. BTW, use ( ) to enclose powers if the powers include 2 terms. For instance, in e, it is not sure if the problem is
3 + 4^2x =5 or 4^(2x+3)=5. THERE IS A BIG DIFFERENCE.

in d. if you take the ln of both sides ,
2x = ln 7. Find what ln 7 is (about 2), and set that equal to 2x; your answer is about 1.

Answer 3

a.) 8^x=(1/32)^x-2
I think you meant (1/32)^x -2 , not 1/32)^(x-2) ? [Did you?]
8^x = (1/32)^x -2
(2^x)^3 - (2^(-5))^x +2 =0
(2^x)^3 - (2^x)^(-5) +2 =0
Set u = 2^x
u^3 - u^(-5) +2 =0
u^8 +2u^3 -1 =0
Solve for u, then for x = log_2 u
u= -1 is an extraneous solution, thus you can extract a
(u+1) factor from u^8 +2u^3 -1

b.) 3^2x -12(3^x) +27=0
(3^x)² -12(3^x) +27=0
Set u = 3^x
u² -12u +27=0
(u-9)(u-3) = 0
u = 3 or 9
3^x = 3 or 9
x = log_3 (u)
=> x = 1 or 2 [SOLUTION]

c.)(4x^3 e^3x) -3x^4e^3x = 0
4x^3 e^3x = 3x^4e^3x
Divide across by the HCF (x^3 e^3x)...
4 = 3x
x =4/3 [SOLUTION]

Answer 4

8^x=(1/32)^x-2
=>2^(3x)=2^(-5)(x-2)
=>3x=-5x+10
=>x=10/8

3^2x-12(3^x)+27=0 let 3^x=a
=>therefore a^2-12a+27=0
=>a=9 or 3
=>x=3 or 1

(4x^3e^3x)-3x^4e^3x=0
taking e^3x common,
e^3x(4x^3-3x^4)=0
(e^3x)(x^3)(4-3x)=0
=>x= -infinity(for e^3x=0) or x=0(for x^3=0) or x=3/4(for 4-3x=0)

e^2x=7
=>2x=ln7
=>x=(1/2)ln7

4^2x+3=5
=>4^2x=2
=>2^4x=2^1
=>4x=1
=>x=1/4

Answer 5

d.) ln (natural log)

ok so take the ln on both sides, ln will cancel out the e. your problem will look like this

2x = ln7
divide the 2.
x = ln7/2.
do that on your calc, make sure to put parans around the ln7.

e.)use log
2x +3 log4 = log 5
divide log4 on both sides.
2x + 3 = log5/ log4
minus 3,
divide by 2.
x = (log5 / log4)-3/2