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# solve and check

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1. 4^2x = 32^1/2

2. 9^x = 3^x-2

3. 2^x = 4^x+1

4. 4^x = 10

5. (1/4)^2x = (1/2)^x

6. 2.4^3x+1 = 9

Apr 25, 2019

#1
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1. 4^2x = 32^1/2  ......the idea is to get the bases the same and then solve for the exponents

(2^2) ^(2x)   = (2^5)^(1/2)

(2)^(4x)  = (2)^(5/2)

4x  = 5/2      divide both sides by 4

x  = 5/8

2. 9^x = 3^x-2

(3^2)^x  =3^(x - 2)

(3)^(2x)  = 3^(x - 2)

2x  = x - 2     subtract x from both sides

x = - 2

3. 2^x = 4^x+1

2^x  = (2^2)^(x + 1)

2^x = (2)^(2x + 2)

x = 2x + 2      subtract 2x from both sides

-1x  = 2         divide both sides by -1

x = -2   Apr 26, 2019
#2
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4.    4^x  = 10

Take the log of each side

log 4^x  = log 10       and we can write

x * log 4  = 1      divide both sides by log 4

x  =  1 / log 4

5. (1/4)^2x = (1/2)^x

[ (1/2)^2 ]^(2x)  = (1/2)^x

(1/2)^(4x)  = (1/2)^x

4x  = x

4x - x  = 0

3x  = 0

x  = 0   Apr 26, 2019
#3
0

Thank you so much, I needed this

Guest Apr 26, 2019
#4
+2

Last one

6.   2.4^(3x+1) = 9    take the log of  both sides

log (2.4) ^(3x + 1)  = log 9      and we can write

(3x + 1) * log 2.4  = log 9

3x * log 2.4 + 1 * log 2.4 =  log 9

(3 log 2.4)x = log 9 - log 2.4     divide both sides by 3 log 2.4

x  =  [ log 9 - log 2.4 ] / [ 3* log 2.4 ] ≈  0.5033   Apr 26, 2019