how do you solve problems like :
log3x-log5=1
log3x-log5=1
log (3x/5) = 1
then 3x/5 = 10^1
3x = 50
x = 50/3
check:
log(3(50/3)) - log5
= log50-log5
= 1.69897 - .69897
= 1
how do u solve
(6/4x^2)+ (2/5x)
To solve the equation log3x - log5 = 1, we need to use properties of logarithms and algebraic manipulations. Here are the steps to solve this problem:
Step 1: Use the property of logarithms that states log(a) - log(b) = log(a/b). Rewrite the equation as a single logarithm:
log (3x/5) = 1
Step 2: Exponentiate both sides of the equation with the base of the logarithm, which in this case is 10. This cancels out the logarithm, giving:
10^(log (3x/5)) = 10^1
Step 3: Simplify the equation:
3x/5 = 10
Step 4: Multiply both sides of the equation by 5 to isolate the variable:
5 * (3x/5) = 10 * 5
3x = 50
Step 5: Divide both sides of the equation by 3:
(3x)/3 = 50/3
x = 50/3
So, the solution to the equation log3x - log5 = 1 is x = 50/3.