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# Algebra

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what is x in each problem ? 1.) 6-log1/2x=3

2.) log (4x-3) +6 = 4

3.) 2/3log5x =2

4.) 2log 250x-6 =4

5.) 5-2logx=1/2

the base for each log is 10

Guest Mar 24, 2017
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#1
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1)

Solve for x:
6 + (log(4 x - 3))/(log(10)) = 4

Subtract 6 from both sides:
(log(4 x - 3))/(log(10)) = -2

Multiply both sides by log(10):
log(4 x - 3) = -2 log(10)

-2 log(10) = log(1/10^2) = log(1/100):
log(4 x - 3) = log(1/100)

Cancel logarithms by taking exp of both sides:
4 x - 3 = 1/100

4 x = 301/100

Divide both sides by 4:

Guest Mar 24, 2017
#3
+6928
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That was question 2.

MaxWong  Mar 24, 2017
#2
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1)$$6 - \log \dfrac{1}{2} x = 3\\ \log \dfrac{1}{2} x = 6 - 3 = 3\\ \dfrac{1}{2}x=10^3\\ x = 2\times 10^3 = 2000$$

2)$$\log(4x-3)+6 = 4\\ \log(4x-3)=-2\\ 4x-3 = 10^{-2}=\dfrac{1}{100}\\ 4x = \dfrac{301}{100}\\ x=\dfrac{301}{400}$$

3)$$\dfrac{2}{3}\log(5x) = 2\\ \log(5x) = 3\\ 5x = 10^3 = 1000\\ x = 200$$

4)$$2\log(250x)-6 = 4\\ 2\log(250x) = 10\\ \log(250x)=5\\ 250x = 10^5 = 100000\\ x = 400$$

5)$$5-2\log x = \dfrac{1}{2}\\ 2\log x = 5-\dfrac{1}{2}=\dfrac{9}{2}\\ \log x = \dfrac{9}{4}\\ x = 10^{9/4}=\sqrt[4]{10^9}=100\cdot\sqrt[4]{10}$$

MaxWong  Mar 24, 2017
#4
+6928
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If my assumptions are wrong, tell me.

MaxWong  Mar 24, 2017

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