logarithms: solve logx-log(x2-2)=1 for x

The exact answer will depend on whether that's the natural logarithm or not. In other words, is it log base e or log base 10? I've seen log(x) used to refer to either of those, so you'd be best served by using your teacher/book's definition.That aside, think about what you know about the log function. What does it represent, and how might you "undo" it? As a hint, consider this: logn(n) = 1.
 
solve for x:

logx-log(x2-2)=1

log(x/x2-2)=1

what next? pls help me

if y is equal to the olg base a of x
y = loga(x)
then x is equal to ay
x = ay

For example, if
log10(x) = 1
then
x = 101 = 10
 
if y is equal to the olg base a of x
y = loga(x)
then x is equal to ay
x = ay

For example, if
log10(x) = 1
then
x = 101 = 10


log(x/x2-2)=1
log(x/x2-2)=log10
(x/x2-2)=10
That makes a quadratic equation in x
solving gor x gives me 1.465

Thanks for your help.
 
The exact answer will depend on whether that's the natural logarithm or not. In other words, is it log base e or log base 10? I've seen log(x) used to refer to either of those, so you'd be best served by using your teacher/book's definition.That aside, think about what you know about the log function. What does it represent, and how might you "undo" it? As a hint, consider this: logn(n) = 1.

Thanks for your help.
 
log(x/x2-2)=1
log(x/x2-2)=log10
(x/x2-2)=10
That makes a quadratic equation in x
solving gor x gives me 1.465

Thanks for your help.

Did you check your answer by putting it back into the original equation?

What happened to the other solution of the quadratic equation?
 
Did you check your answer by putting it back into the original equation?

What happened to the other solution of the quadratic equation?

other solution is negative that´s why discarded.
I didn´t chk my answer by substituting. I checked in answer (solution) o book
 
other solution is negative that´s why discarded.
I didn´t chk my answer by substituting. I checked in answer (solution) o book

Is there a physical reason for discarding negative solution?
 
when i put x negative in original question (just to check my question) it will make log negative as log(-)-log(.....) not defined

That is correct!!

We need to be careful about the original equation.
 
log(x/x2-2)=1
log(x/x2-2)=log10
(x/x2-2)=10

newuser, you must put those denominators in grouping symbols:

log[x/(x2 - 2)] = 1
log[x/(x2 - 2)] = log10
x/(x2 - 2) = 10


-- - - -- - - - -- - - - - - - - - - - - - - - - - - - - - - - - -------


logarithms can´t be negative


That is not true. Logarithms can be negative, zero, or positive. They are exponents.

But you cannot take the logarithm of zero or of a negative number to get a real number result.



Sometimes a negative solution is retained, and the positive solution is rejected after checking.

Or, there might be no solution.


 
Last edited:
newuser, you must put those denominators in grouping symbols:

log[x/(x2 - 2)] = 1
log[x/(x2 - 2)] = log10
x/(x2 - 2) = 10


-- - - -- - - - -- - - - - - - - - - - - - - - - - - - - - - - - -------





That is not true. Logarithms can be negative, zero, or positive. They are exponents.

But you cannot take the logarithm of zero or of a negative number to get a real number result.



Sometimes a negative solution is retained, and the positive solution is rejected after checking.

Or, there might be no solution.


Yes, you are correct but I answered according to the original question that I have.
Thanks
 
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