solve
-4+7logx=10
&
log(x+3)+log(x-2)=log14
(1) -4 + 7log(x) = 10
7log(x) = 14
log(x) = 2
x = 10^2 = 100
(2) log(x+3) + log(x-2) = log 14
recall some laws of exponents. when we add two log terms with same base, we can combine them by multiplying the terms inside the log. thus, we can rewrite this as,
log[(x+3)(x-2)] = log 14
equating the terms inside,
(x+3)(x-2) = 14
x^2 + x - 6 = 14
x^2 + x - 20 = 0
since it's factorable,
(x+5)(x-4) = 0
x = -5 and x = 4
but note that x = -5 is extraneous since if this substituted back to original,
log(-5+3) + log(-5-2)
log(-2) + log(-7)
..and log of any negative number is undefined, thus
x = 4 only.
hope this helps~ :)
To solve the equations, we will follow these steps:
Equation 1: -4 + 7log(x) = 10
Step 1: Move the constant term to the right side:
7log(x) = 10 + 4
Step 2: Simplify the equation:
7log(x) = 14
Step 3: Isolate the logarithmic term by dividing both sides by 7:
log(x) = 14/7
Step 4: Simplify the right side:
log(x) = 2
Step 5: Rewrite the equation in exponential form:
x = 10^2
Step 6: Solve for x:
x = 100
Therefore, the solution to the first equation is x = 100.
Equation 2: log(x+3) + log(x-2) = log(14)
Step 1: Apply the product rule of logarithms on the left side:
log((x + 3)(x - 2)) = log(14)
Step 2: Remove the logarithms by equating the expressions inside them:
(x + 3)(x - 2) = 14
Step 3: Expand and simplify the equation:
x^2 + x - 6 = 14
Step 4: Move the constant term to the right side:
x^2 + x - 6 - 14 = 0
Step 5: Simplify further:
x^2 + x - 20 = 0
Step 6: Factor the quadratic equation:
(x - 4)(x + 5) = 0
Step 7: Set each factor to zero and solve for x:
x - 4 = 0 --> x = 4
x + 5 = 0 --> x = -5
Therefore, the solutions to the second equation are x = 4 and x = -5.