Use the result of the first paragraph to change $\log_2(2x+1)$ to $2\log_4(2x+1)$ and $\log_2x$ to $2\log_4x$; then you have ... and you can use the usual properties of logs to express this as the log base $4$ of a single expression. More on math.stackexchange.com

1+2*(log base 4 of (x+1))=2*(log base 2 of (x)) (Given) Writing everything in terms of log base 4 is an acceptable strategy. Let's start with the 1 on the left-hand-side. 1 How do we re-write the 1, so that it's in terms of log base 4? Well, let's find out. Iog base 4 of (y) = 1 Suppose we didn't know what we need to take log base 4 of, in order to get 1. Let's call this value y, as shown above. We can convert this equation from logarithmic form to exponential form. Doing this gives us that 41=y. Then we know that 41=4, so we get 4=y (So, y=4). We can re-write 1 as (log base 4 of (4)) (I will call this expression (*)). Let's re-write the log on the right-hand-side as a log with a base of 4. 2*(log base 2 of (x)) We have a coefficient of 2 on this log, so we first need to pull the 2 back inside of the log. When we do this, the 2 will become the exponent on the x. In this case, we are just using the Power Property of Logarithms in the opposite direction. (log base 2 of (x2)) We can re-write this expression as a log with a base of 4, by using the Change-of-Base Formula. ((log base 4 of (x2))/(log base 4 of (2))) We can evaluate (log base 4 of (2)) by hand. Let's suppose that we didn't know the value of (log base 4 of (2)) off the top of our heads, so let's call it z. (log base 4 of (2))=z Converting from logarithmic form to exponential form gives us that 4z=2. We can get common bases, since we can think of 4 as 22 (and the 2 is 21). Then, using our properties of exponents, since we have a power raised to a power, we leave the base alone, and then multiply the exponents together. So, 4z=(22)z=22*z=22z. We can think of the 2 on the right as 21. We now have the following: 22z=21 At this point, since we just have a single exponential expression on both sides, and both sides have the same base, we can just set the exponents equal to each other, and solve (This is because exponential functions are one-to-one). 2z=1 z=(1/2) (Divide both sides by 2) Remember from earlier that z was what we called our original log (i.e., log base 4 of (2)), so this tells us that (log base 4 of (2))=(1/2). ((log base 4 of (x2))/(log base 4 of (2))) ((log base 4 of (x2))/(1/2)) Dividing by (1/2) is the same thing as multiplying by (2/1) (which is 2). 2*(log base 4 of (x2)) Now we pull the 2 back inside of the log, and when we do this, it becomes the exponent on the x2. (log base 4 of ((x2)2)) (x2)2=x2*2=x4 (log base 4 of (x4)) (I will call this expression (***)) As for the 2nd log on the left-hand-side, we have the following: 2*(log base 4 of (x+1)) Once again, we pull the 2 back inside of the log, and when we do this, it becomes the exponent on the (x+1). (log base 4 of ((x+1)2)) (log base 4 of (x2+2x+1)) (I will call this expression (**)) I will expand the (x+1)2. (x+1)2=((x)2)+(2*(x)*(1))+((1)2)=x2+2x+1 ((x)2)=x2 2*(x)*(1) =2x*(1) (Because 2*(x)=2x) =2x (Because 2x*(1)=2x) (1)2=12=1*1=1 Note: I used a formula to expand (x+1)2. Since we have a binomial sum/difference being squared, we can use this formula to expand the expression. (a±b)2=a2±2ab+b2 Note: If you're having trouble visualizing the multiplication this way, then you could also re-write (x+1)2 as (x+1)*(x+1), and then multiply/F.O.I.L. it out. Just remember that exponents do NOT distribute across sums and differences, so if you have a binomial sum/difference, you can't just square each term individually, and then add/subtract their squares. Our equation now becomes as follows: () + () = () (log base 4 of (4))+(log base 4 of (x2+2x+1))=(log base 4 of (x4)) On the left-hand-side, we just have two logs with the same base that are being added, so we can write it in terms of just a single logarithm (Of course, the new log will have the same base as the two logs that we're adding), and when we do this, we multiply the arguments of these logs together (This will give us the argument of the condensed log). (log base 4 of (4))+(log base 4 of (x2+2x+1)) =(log base 4 of (4*(x2+2x+1))) Keep in mind that the (x2+2x+1) needs to be inside parentheses, because that whole expression (not just one single term, but every term) is being multiplied by the 4. Now, we distribute the 4. =(log base 4 of (4x2+8x+4)) The right-hand-side has not changed. (log base 4 of (4x2+8x+4))=(log base 4 of (x4)) We now have a logarithmic equation, where we just have a single log on both sides, and both logs have the same base, so we can just set the arguments (i.e., the insides of the logs) equal to each other, and then solve for x (Similar to exponential functions, logarithmic functions are also one-to-one). 4x2+8x+4=x4 This is a polynomial equation of degree higher than 1, so we need to get one side of the equation equal to zero (0). In the equation above, the term with the highest power is x4, so I will move everything over to the right-hand-side (I don't like to have a negative leading coefficient). Hence, we subtract 4x2, 8x, and 4 from both sides. We get the following: 0=x4-4x2-8x-4 I will re-write the equation, so that we have the variables on the left-hand-side. x4-4x2-8x-4=0 You will need to solve this equation for x, and then check the original equation for extraneous solutions. More on reddit.com

Many reasons. One is that logs are an invertible function. If you allow for log base -2 and log base 2 of 4 to both be 2, you're going to lose invertibility. Also because of change of base. Log base -2 of 4 should be equal to log(4) / log(-2) but we've got a problem with that denominator. More on reddit.com

The logarithm of negative numbers is undefined for this reason. No power of 2, or any number, will ever yield a negative. This is why the graph of log(x) has a vertical asymptote at x=0 as it is undefined in the negatives. More on reddit.com