log4+log(x-7)=2
remember loga-logb=log(a/b)
Do that, then take the antilog of each side, solve for x
the antilog of 2 is 100, as in 10^2=100
But the question above is loga+logb=loga*b
same principle
log(4(x-7))=2
4(x-7)=100
solve for x.
Okay, thanks :)
To solve the equation log4 + log(x-7) = 2, we can use logarithmic properties.
1. Combine the logarithms on the left-hand side of the equation using the product rule of logarithms which states that log(a) + log(b) = log(ab):
log4 + log(x-7) = log(4(x-7))
2. Now, the equation becomes log(4(x-7)) = 2.
3. Convert the logarithmic equation into its exponential form. For any positive base "b" and real numbers "y" and "x", if log_b(x) = y, then b^y = x. In this case, the base is 10 (since it's a common logarithm) and the exponent is 2:
10^2 = 4(x-7).
4. Simplify the equation by solving for x:
100 = 4x - 28.
5. Move the variable terms to one side of the equation:
4x = 100 + 28
4x = 128.
6. Divide both sides of the equation by 4 to isolate x:
x = 128/4
x = 32.
Thus, the solution to the equation log4 + log(x-7) = 2 is x = 32.