Solve : 2^X * 2^(X + 1) =128?

Question

Solve the equation:

This is what I did

X LOG2 * (X+1)LOG2 = LOG128

Then ??

Answer 1

x = 3

Careful with that first step. You can do it that way, but remember that log (mn) = log m + log n, not log m * log n:
2^x * 2^(x + 1) = 128
log (2^x * 2^(x + 1)) = log 128
log (2^x) + log(2^(x + 1)) = log 128
x log 2 + (x + 1) log 2 = log 128
Divide both sides by log 2:
x + (x + 1) = log 128 / log 2 = log (base 2) 128 = 7
2x + 1 = 7
x = 3

It's probably easier to do the same thing more like this, though:
2^x * 2^(x + 1) = 128
2^(x + (x + 1)) = 128
2^(2x + 1) = 128
2^(2x + 1) = 2^7
2x + 1 = 7
x = 3

Answer 2

Perhaps this is easier.
128 is a power of 2. i.e. 128 = 2^7.
This problem is illustrating the rules of powers and bases.
If the base is the same, add the exponents.
2^x * 2^(x+1) is the same as 2^(x+x+1).
So we have 2^(2x+1) = 2^7.
The bases are the same on each side, so that means we can set the exponents equal to each other:
2x+1 = 7. Now solve for x:
x = 3

Answer 3

x = 3

Answer 4

You're distributing the logs incorrectly. When you take the log of both sides, you get
Log [ (2^x) * 2^(x+1) ] = log(128)

You could then change this to
Log [ (2^x) ] + Log [ 2^(x+1) ] = log(128)
x log(2) + (x+1)log(2) = log(128)
x + (x+1) = log(128) / log(2)
2x + 1 = log(128) / log(2)
x = ([log(128) / log(2)] - 1 ) /2

Notice though that you can write the log part as:
log [base 2] (128) =
log [base 2] (2^7) =
7 log [base 2] (2) =
7
So x = (7-1)/2 = 6/2 = 3.

Another way to go about this is:
(2^x) 2^(x+1) = 128
2^(x + x+1) = 128
2^(2x + 1) = 2^7
So 2x + 1 = 7, etc.

Answer 5

2^(2x + 1) = 128
2x + 1 = 7
x = 3

Answer 6

hey............dont use log. y u r going complicated
just split them.........................
2^x *2^x *2=128 (remember m^(n+k)=m^n*m^k..)
let 2^x =m
m*m*2=128
m^2=64
m=8
2^x=8
2^x=2^3
equating them
x=3

Answer 7

quit the log.
take the advice of doing the simpler way. why complicate it?

Answer 8

x=29 try it yourself i am positive on this one i have never been more sure of anything in my life.