here is the question:
log5(x-4)= log7x solve for x.
These are just base 10 logs.
log100 = 2
This equation has the same format as log 40 = log (2x20)
Since both sides have log base 10, you divide by log base 10 and end up with 5(x-4) = 7x, so 5x -20 =7x x=-10
One doesn't exactly "divide by base 10".
log5(x-4)= log7x solve for x.
take each side as exponents to the base 10, or
10^(log5(x-4))= 10^(log7x)
then
5(x-4) = 7x
and x= 10
check...
log (5(-14)= log -70
log (-70)=log(-70) and it checks
Did not see the answer.
Log5(x-4)=7x. Now remove bracket 5x-20=7x then take 5x on the other side=-20=7x-5x =-20=2x which is x=_10
Yes, that is correct!
To solve the equation log5(x-4)= log7x for x, you can follow these steps:
1. Start by converting the logarithmic equation into exponential form. For base 10 logarithms, the exponential form is 10^(log10(x)) = x. In this case, we have:
10^(log5(x-4)) = 10^(log7x)
2. Simplify the equation using the properties of exponents. The exponents can cancel out with the base 10 logarithms:
x - 4 = x
3. Since we have x on both sides of the equation, subtract x from both sides to isolate the variable:
x - x - 4 = 0
-4 = 0
4. This is a contradiction. Since -4 is not equal to 0, there is no solution to the equation.