Solving Log Equation

Question

log(4)+log(x)=log(5)-log(x) (SOLVE)

I understand the log(x) is to be a ten and in putting the 4 and five log into the calculator I got extremely long decimals:

log(4)= .6020599913

log(5)= .6989700043

log(10)= 1

log(4)+log(10)=log(5)-log(10)

1.602059991= -0.3010299957

I am not sure if this is the right method and if this is correct so far...what exactly is there to solve?

Answer 1

The standard forms of a logarithmic function are these, x = b^y which is written y = log_b(x). When the b (base) is 10 it's not indicated but implied instead. So we can't assume that x=10 here because it's not the base. The way to solve this is through log laws. 

We must first separate the two types of logs we have, ones with the variable and the others with the constants. By doing this we get:

2log(x) = log(5)-log(4)

By log laws, log(x)-log(y)=log(x/y) thus we get log(5/4) on the right side. 

Next we divide by 2 on each side to get:

log(x)=1/2log(5/4). Again, by log laws the 1/2 becomes the exponent to the 5/4 inside the log function:

log(x)=log((5/4)^(1/2)). From here we can remove the log on both sides because for both sides to equal each other, we just need x to equal (5/4)^(1/2). And now the final steps are just to clean up the right side:

x=√(5/4)=√5/√4=√5/2 and that's our answer. 

Hope this helped! =]

Answer 2

log(4)+log(x) = log(5)-log(x)

Collect variables in one side, and constants in the other side,

2log(x) = log(5/4)

log(x) = (1/2)log(5/4)

log(x) = log(sqrt(5/4))

Answer: x = sqrt(5/4) = sqrt(5)/2