SOLUTION: Hi, could you help me to solve these equations? 2^(2x+1) - 10.2^x=-12 log(4,[log(4,(log4,x))])=0 1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Hi, could you help me to solve these equations? 2^(2x+1) - 10.2^x=-12 log(4,[log(4,(log4,x))])=0 1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))       Log On


   

Question 202660: Hi, could you help me to solve these equations?
2^(2x+1) - 10.2^x=-12
log(4,[log(4,(log4,x))])=0
1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to give you an idea on how to tackle these problems.


2%5E%282x%2B1%29-10%2A2%5Ex=-12 Start with the given equation.


2%5E%282x%29%2A2%5E1-10%2A2%5Ex=-12 Break up the exponent using the identity x%5E%28y%2Bz%29=x%5E%28y%29%2Ax%5E%28z%29


2%5E%282x%29%2A2-10%2A2%5Ex=-12 Raise 2 to the first power to get 2


2%2A2%5E%282x%29-10%2A2%5Ex=-12 Rearrange the terms.


2%2A%282%5Ex%29%5E2-10%2A2%5Ex=-12 Rewrite 2%5E%282x%29 as %282%5Ex%29%5E2 using the identity %28x%5E%28y%29%29%5Ez=x%5E%28y%2Az%29


Now let z=2%5Ex. So z%5E2=%282%5Ex%29%5E2


2z%5E2-10z=-12 Replace %282%5Ex%29%5E2 with z%5E2. Replace 2%5Ex with z


2z%5E2-10z%2B12=0 Add 12 to both sides.


Notice that the quadratic 2z%5E2-10z%2B12 is in the form of Az%5E2%2BBz%2BC where A=2, B=-10, and C=12


Let's use the quadratic formula to solve for "z":


z+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


z+=+%28-%28-10%29+%2B-+sqrt%28+%28-10%29%5E2-4%282%29%2812%29+%29%29%2F%282%282%29%29 Plug in A=2, B=-10, and C=12


z+=+%2810+%2B-+sqrt%28+%28-10%29%5E2-4%282%29%2812%29+%29%29%2F%282%282%29%29 Negate -10 to get 10.


z+=+%2810+%2B-+sqrt%28+100-4%282%29%2812%29+%29%29%2F%282%282%29%29 Square -10 to get 100.


z+=+%2810+%2B-+sqrt%28+100-96+%29%29%2F%282%282%29%29 Multiply 4%282%29%2812%29 to get 96


z+=+%2810+%2B-+sqrt%28+4+%29%29%2F%282%282%29%29 Subtract 96 from 100 to get 4


z+=+%2810+%2B-+sqrt%28+4+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


z+=+%2810+%2B-+2%29%2F%284%29 Take the square root of 4 to get 2.


z+=+%2810+%2B+2%29%2F%284%29 or z+=+%2810+-+2%29%2F%284%29 Break up the expression.


z+=+%2812%29%2F%284%29 or z+=++%288%29%2F%284%29 Combine like terms.


z+=+3 or z+=+2 Simplify.


So the solutions (in terms of 'z') are z+=+3 or z+=+2


Now recall that we let z=2%5Ex. So let's use this to find the solutions in terms of "x".


z+=+3 Start with the first solution in terms of 'z'


2%5Ex+=+3 Plug in z=2%5Ex


log%2810%2C%282%5Ex%29%29+=+log%2810%2C%283%29%29 Take the log of both sides


x%2Alog%2810%2C%282%29%29+=+log%2810%2C%283%29%29 Rewrite the first log using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29


x+=+log%2810%2C%283%29%29%2Flog%2810%2C%282%29%29 Divide both sides by log%2810%2C%282%29%29 to isolate "x"


x+=+log%282%2C%283%29%29 Use the change of base formula to simplify


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z+=+2 Start with the second solution in terms of 'z'


2%5Ex+=+2 Plug in z=2%5Ex


log%2810%2C%282%5Ex%29%29+=+log%2810%2C%282%29%29 Take the log of both sides


x%2Alog%2810%2C%282%29%29+=+log%2810%2C%282%29%29 Rewrite the first log using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29


x+=+log%2810%2C%282%29%29%2Flog%2810%2C%282%29%29 Divide both sides by log%2810%2C%282%29%29 to isolate "x"


x+=+log%282%2C%282%29%29 Use the change of base formula to simplify


x+=+1 Evaluate the log base 2 of 2 to get 1


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Answer:


So the solutions are x+=+log%282%2C%283%29%29 or x+=+1