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klasskru [66]
4 years ago
10

Help me..... 25^x=5^x-2

Mathematics
2 answers:
QveST [7]4 years ago
7 0
Ok, under the assumption that the problem is
25^x = 5^{x-2}

as opposed to
25^x = 5^x-2


25^x = 5^{x-2}
looking the the left side

25 = 5^2 so the left side can be written as
(5^2)^x = 5^{x-2}
now, if you remember exponent rules, a power to a power, you just multiply them together
so it would be equal to this equation
5^{2x} = 5^{x-2}

now, since they share the same base (5)
you can jsut compare the exponents

2x= x-2

and then solve for x

trying to solve
25^x = 5^x-2 
requires the use of log i believe

Vikentia [17]4 years ago
4 0
Solve for x right ? . if u find for x . the answer ís x=-2
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6 0
1 year ago
Read 2 more answers
Select the correct answer from each drop-down menu.
mr_godi [17]

Answer:

correct option for first blank is 5/4 and for second blank is \frac{ 3i\sqrt{7}}{4}

i.e m= \frac{5}{4}\pm\frac{ 3i\sqrt{7}}{4}

Step-by-step explanation:

The given equation

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We can use quadratic formula to solve this question.

The above equation can be written as: m^2 - \frac {5m}{2} + \frac{11}{2} = 0

and the formula used will be:

m= \frac{-b\pm\sqrt{b^2-4ac}}{2a}

Putting values of a= 1, b= -5/2 and c= 11/2 and solving we get:m=\frac{-\frac{-5}{2}\pm\sqrt{{(\frac{-5}{2})}^2-4(1)(\frac{11}{2})}}{2(1)}\\\\m=\frac{\frac{5}{2}\pm\sqrt{(\frac{25}{4})-22}}{2}\\m=\frac{\frac{5}{2}\pm\sqrt{(\frac{-63}{4})}}{2}\\m= \frac{\frac{5}{2}}{2}\pm\frac{\sqrt{(\frac{-63}{4})}}{2}\\m= \frac{5}{4}\pm\frac{\sqrt{-63}}{4}

Since there is - sign inside the √ so \sqrt{-1} is equal to i and we have to divide \sqrt{63} into its multiples such that the square root of one multiple is whole no so,

\sqrt{63}  = \sqrt{9}* \sqrt{7}=3* \sqrt{7}

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the value of m= \frac{5}{4}\pm\frac{ 3i\sqrt{7}}{4}

so, correct option for first blank is 5/4 and for second blank is \frac{ 3i\sqrt{7}}{4} .

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