I can't get this, please help with steps, thank you.
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Indices and logs are linked together and sometimes using logs really simplifies the question instead of using the rules of indices to solve for x
Just like all maths questions whatever you do to one side of the equation you also have to do to the other side of the equation
So the easiest way to do this question is to get the log of each side
log 5^(8 x − 8) = log 8^(3 x − 3)
One of the rules of logs is: log a^b = b(log a)
you can use this rule here to make the question alot easier
(8x-8)log5 = (3x-3)log8
mutiply out the brackets
8xlog5 -8log5 = 3xlog8 -3log8
bring everything with an x to the left hand side and everything else to the right hand side
8xlog5 - 3xlog8 = -3log8 + 8log5
take out whats common which is x
x(8log5 -3log8) = -3log8 + 8log5
divide both sides by (8log5 -3log8)
x = -3log8 + 8log5 / (8log5 -3log8)
-3log8 + 8log5 / ( -3log8 + 8log5) = 1
x = 1
Just like all maths questions whatever you do to one side of the equation you also have to do to the other side of the equation
So the easiest way to do this question is to get the log of each side
log 5^(8 x − 8) = log 8^(3 x − 3)
One of the rules of logs is: log a^b = b(log a)
you can use this rule here to make the question alot easier
(8x-8)log5 = (3x-3)log8
mutiply out the brackets
8xlog5 -8log5 = 3xlog8 -3log8
bring everything with an x to the left hand side and everything else to the right hand side
8xlog5 - 3xlog8 = -3log8 + 8log5
take out whats common which is x
x(8log5 -3log8) = -3log8 + 8log5
divide both sides by (8log5 -3log8)
x = -3log8 + 8log5 / (8log5 -3log8)
-3log8 + 8log5 / ( -3log8 + 8log5) = 1
x = 1
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In general, if one wants to solve an equation in which the variable occurs in exponents, one must take the logarithm of both sides of the equation. (If the variable occurs both in exponents and not in exponents, one is stuck with approximating solutions, but you have an equation where it only occurs in exponents, so there is hope of getting an exact solution by algebraic means.)
What logarithm? Any logarithm will do. Personally I like natural logarithms, so we get
ln(5^(8x - 8) ) = ln(3^(3x - 3) )
Now by laws of logarithms, this simplifies to
(8x - 8) * ln(5) = (3x - 3) * ln(3)
(Notice why any logarithm function works: they all satisfy the law log(a^b) = b * log(a), so any
of them will get the variables out of the exponent.)
This is now a linear equation (albeit with the ugly numbers ln(5) and ln(3) among its coefficients),
so distribute the factors, move the terms with x's to one side, the ones without to the other and divide.
You'll get 1, the same as the computer told you.
What logarithm? Any logarithm will do. Personally I like natural logarithms, so we get
ln(5^(8x - 8) ) = ln(3^(3x - 3) )
Now by laws of logarithms, this simplifies to
(8x - 8) * ln(5) = (3x - 3) * ln(3)
(Notice why any logarithm function works: they all satisfy the law log(a^b) = b * log(a), so any
of them will get the variables out of the exponent.)
This is now a linear equation (albeit with the ugly numbers ln(5) and ln(3) among its coefficients),
so distribute the factors, move the terms with x's to one side, the ones without to the other and divide.
You'll get 1, the same as the computer told you.
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ln(5^(8x-8))=ln(8^(3x-3))
use log rules. ln(a^b) = b*ln(a)
(8x-8)ln5 = (3x-3)ln8
x(8ln5-3ln8) = 8ln5-3ln8
x = (8ln5-3ln8) / (8ln5-3ln8)
x=1
check!
5^0=8^0
1=1
make it a good day
use log rules. ln(a^b) = b*ln(a)
(8x-8)ln5 = (3x-3)ln8
x(8ln5-3ln8) = 8ln5-3ln8
x = (8ln5-3ln8) / (8ln5-3ln8)
x=1
check!
5^0=8^0
1=1
make it a good day
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