math help, no this isnt homework?

Question

here are three problems, that may involve logarithims, please give me anything useful:
1. 4^x-2*5^(2x)<10^x
2. 7*3^(x+1)-5^(x+2)=3^(x+4)-5^(x+3)
3. 6*9^(1/x)-13*6^(1/x)+6*4^(1/x)=0

the second one finishes with -5^(x+3), it was cut off, and the last one finishes with 6*4^(1/x)=0

Answer 1

Question 1 : 4^x - 2 * 5^(2x) < 10^x

Factorise :
2^(2x) - 2 * 5^(2x) < 2^x * 5^x

Divide through by 2^(2x) :
1 - 2 * (5/2)^(2x) < (5/2)^x

Let y = (5/2)^x

Therefore, 1 - 2y^2 < y
or, 2y^2 + y - 1 > 0
or, (y + 1)(2y - 1) > 0
Thus, y < -1 or y > 1/2

That is, (5/2)^x < - 1 or (5/2)^x > 1/2

But (5/2)^x is always > 0, so the only
solution occurs when (5/2)^x > 1/2.

Take the reciprocal of both sides :
(2/5)^x < 2

Take logarithm of both sides :
x(log2 - log5) < log2

Therefore, x < log2 / (log2 - log5)
or x < -0.7565 approximately.
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Question 2 : 7 * 3^(x + 1) - 5^(x + 2) = 3^(x + 4) - 5^(x + 3)

Divide through by 3^(x + 1) :
7 - 5 * (5/3)^(x + 1) = 27 - 25 * (5/3)^(x + 1)

Rearrange :
20 * (5/3)^(x + 1) = 20

Therefore, (5/3)^(x + 1) = 1 = (5/3)^0
Thus, x + 1 = 0, so, x = -1
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Question 3 : 6 * 9^(1/x) - 13 * 6^(1/x) + 6 * 4^(1/x) = 0

Factorise for ease of viewing :
6 * 3^(2/x) -13 * 2^(1/x) * 3^(1/x) + 6 * 2^(2/x) = 0

Divide through by 6^(1/x), which equals 2^(1/x) * 3^(1/x) :
6 * (3/2)^(1/x) - 13 + 6 * (2/3)^(1/x) = 0

The latter term, (2/3)^(1/x) = 1 / [ (3/2)^(1/x) ]

So now we have : 6 * (3/2)^(1/x) - 13 + 6 / [ (3/2)^(1/x) ] = 0

Let y = (3/2)^(1/x)

Therefore, 6y - 13 + 6 / y = 0

Multiply through by y :
6y^2 - 13y + 6 = 0
or, (3y - 2)(2y - 3) = 0
or, y = 2/3 or 3/2

That is, (3/2)^(1/x) = 2/3 or (3/2)^(1/x) = 3/2

First, trying (3/2)^(1/x) = 2/3 = 1 / (3/2) = (3/2)^(-1)
Therefore, 1 / x = -1, so x = -1.

Second, trying (3/2)^(1/x) = 3/2 = (3/2)^1
Therefore, 1 / x = 1, so x = 1.

So there are 2 solutions, x = -1 or 1.

Answer 2

Im thinking of your question, its not that easy.

I tried this:

4^x/10^x - 2* 5^(2x)/10^x < 10^x/10^x

(2/5)^x - 2*25^x/10^x <1
(2/5)^x - 2*(5/2)^x <1

(2/5)^x - 2*(2/5)^-x < 1

(2/5)^x [1 - 2*(2/5)^(-2x)] < 1

1 - 2 * [(5/2) ^x]^2 < (5/2)^x

Then you can change variable and solve this a second grad equation

(5/2)^x = z


Hope this helps

Ana

Answer 3

carry on with the order of PEMDAS (Parentheses, Exponents, Multiplication, branch, Addition, Subtract) [ (ninety + 36 - 4) ÷ 2 ] x 12 [(122) ÷ 2] x 12 sixty one X 12 = 732 desire I helped :)

Answer 4

2. 7*3^(x+1) - 5^(x+2) = 3^(x+4) - 5^(x+3)
7*3^(x+1)-3^(x+4) = 5^(x+2) -5^(x+3)
3^{x+1} ( 7 - 3^3) = 5^{x+2}( 1- 5)
3^{x+1} ( -20) = 5^{x+2}( -4)
3^{x+1} ( 5) = 5^{x+2}
ln( 3^{x+1} ( 5) ) = ln( 5^{x+2})
ln ( 3^{x+1}) + ln(5) = ln( 5^{x+2})
(x+1) ln(3) + ln(5) = (x+2) ln(5)
(x+1) ln(3) + (x+2) ln(5) = - ln(5)
x( ln(3) + ln(5) ) + ln(3) +2ln(5) = -ln(5)
x( ln(3) + ln(5) ) = - ln(3) -2ln(5) -ln(5)
x = ( -ln(3) -3ln(5) ) / ( ln(3) + ln(5) )


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