math help???????

Question

9^(x+1)=27^(x-1)
is there any other way to solve this equation without using Equating Exponents method?

thanks for your answer but that is not what i mean.
what i mean is without changing the 27 to 3^3

Answer 1

Take the natural log of both sides

ln 9^(x+1) = ln 27^(x-1)

(x+1) ln 9 = (x-1) ln 27

Distribute

ln 9 * x + ln 9 = ln 27 * x - ln 27

ln 9 + ln 27 = ln 27 * x - ln 9 * x = x ( ln 27 - ln 9)

x = (ln 9 + ln 27) / (ln 27 - ln 9) = ln (9*27) / ln (27/9)
= ln 243 / ln 3 = 5

Answer 2

Actually, the equating methods is the better one as opposed to the alternative. Here's the alternative:

Take the log[base 9] of both sides

log[base 9](9^(x+1)) = log[base 9](27^(x - 1))

Recall that whenever we have a power inside a log, we can move that power outside the log as a non-power.

(x + 1)log[base 9](9) = (x - 1) log[base 9](27)

Note that log[base 9](9) = 1, so we can simplify this.

(x + 1) = (x - 1) log[base 9](27)

Now, expand the right hand side.

x + 1 = x log[base 9](27) - log[base 9](27)

Bring all x-terms to the left hand side

x - xlog[base 9](27) = -1 - log[base 9](27)

Factor the x out of the LHS:

x (1 - log[base 9](27)) = -1 - log[base 9](27)

Divide both sides by 1 - log[base 9](27) to isolate the x.

x = [-1 - log[base 9](27)] / [1 - log[base 9](27)]

At this point, we individualy solve for log[base 9](27) by using the change of base formula (this equals ln(27) / ln(9) ). We're going to end up with a messy decimal that may or may not work out due to rounding errors.

Why do that when you can just equate the exponents? It's actually a GOOD thing when we can equate the exponents because we then get a nice clean answer.

9^(x + 1) = 27^(x - 1)

Note that 9 = 3^2 and 27 = 3^3. Replacing, we get

(3^2)^(x + 1) = (3^3)^(x - 1)

Whenever we have a power to a power, we can multiply the exponents.

3^(2[x + 1]) = 3^(3[x - 1])

Now, we expand each exponent.

3^(2x + 1) = 3^(3x - 3)

Since we now have the same base, we can equate the exponents.

2x + 1 = 3x - 3

Solving as normal, we should end up with x = 4.

*****
The reason why we're lucky to even be able to equate exponents is for situations like these:

2^x = 3

Here, it's not possible to equate the exponents. What you have to do is change this to logarithmic form:

log[base 2](3) = x

Which you would then use the change of base formula to get

x = ln(3) / ln(2)

Which works out to be a decimal number best punched into your calculator.

The point I'm trying to make: when you can equate exponents, DO IT! Don't try to look for alternative methods, especially when it's so much less hectic.

Answer 3

9^(x+1)=27^(x-1)
Changing the basis for 3:
9 = 3*3 = 3²
27 = 3*3*3 = 3³
So,
3²^(x + 1) = 3³^(x - 1)
Calc the exponents:
2(x+1) = 3(x-1)
2x + 2 = 3x - 3
2x - 3x = -3 - 2
-x = -5
x = 5
Prove:
9^(x+1)=27^(x-1)
9^(5+1)=27^(5-1)
9^(6) = 27^(4)
3²*6 = 3³*4
3¹² = 3¹²
The answer is 5.
<>><

Answer 4

Yes there is a way

Frist you need to set the 9 and 27 have the same exponential principal
9=3^2 , 27=3^3
*Then you can plug this in
9^(x+1)=27^(x-1) -> 3^2(x+1)=3^3(x-1)

Since both exponents are the same you can say that:

2(x+1)=3(x-1)

Use distributive properties then solve

2x+2=3x-3 therefore x=5

Answer 5

it is better if you would transpose it into logarithmic form:

log9(27^(x-1)) = x+1 :9 is subscript
then it would be (x - 1) log9 (27) = x+1
then find the log value of Log9(27)
value is 1.5 or 3/2
substitute to the equation
: (x-1)(3/2) = x+1
(3x/2) - (3/2) = (x+1)

combine same variables through transposition

(3x/2) - x = (3/2) + 1

((3x-2x)/2) = 5/2

x/2 = 5/2

then multiply both sides by 2

2((x/2) = (5/2))2

therefore 2 would be cancelled and you would get

x = 5

checking:

9^ (5+1) = 27^ (5-1)

9^6 = 27^(4)

531441 = 531441
so thats it!

Answer 6

Well you can always use logs (no not the wooden ones!) but those are more complicated to most than EE.

EE is probably the easiest way to solve this problem. First we need a common base (the number under the exponent), 3 looks good to me. Let's do some math:

9^(x+1)
=(3^2)^(x+1)
=3^(2(x+1)) >>> Power to a power is multiplication.

27^(x-1)
=(3^3)^(x-1)
=3^(3(x-1)) >>> again power to a power.

Therefore we have: 3^(2x+2)=3^(3x-3)

Now since they both have the same base the exponents are equivalent:

2x+2 = 3x-3

Now just solve for x.

Answer 7

logarithms help for this:
9^(x+1) = 27 ^ (x-1)

log of both sides:
log[9^(x+1)] = log[27 ^ (x-1)]

Exponents in a logarithm:
(x+1) * log 9 = (x-1) * log 27

(x+1) * log 3^2 = (x-1) * log 3^3

2(x+1) * log 3 = 3(x-1) * log 3

2 * (x + 1) = 3 * (x - 1)

5 = x

Check

9 ^6 = 27^4

Answer 8

Use logs

2(x+1) ln 3=3(x-1) ln 3 divide both sides by ln 3
2x+2=3x-3 add -2x+3 to each side
5=x
x=5

Answer 9

dividingtheRHS by LHS
27^(x-1)/9^(x+1)=1
3^3(x-1)/3^2(x+1)=1
3^3x-3-2x-2=1
3^x-5=3^0
x-5=0
adding 5
x=5

Answer 10

i'm not sure what "equating exponents" method is.
I'd boil the equation down to 3^[2(x+1)] = 3^[3(x-1)], is that what you're talking about?