In order to solve for x, we must first rewrite the log in exponential form. 

Every log is written in the below general form: 

\(\displaystyle log_{base}(result)=exponent\ is\ rewritten\ as: base^e^x^p^o^n^e^n^t=result\)

In this case we have: 

\(\displaystyle log_{x}(49)=2\)

This becomes: 

\(\displaystyle x^2=49\)

We can solve this by taking the square root of both sides: 

\(\displaystyle \sqrt{x^2}=\sqrt{49}\)

\(\displaystyle x=\pm7\)

Use rules of logarithms...

Take the base of the log and raise it to the number on the right side of the equal sign (which becomes the exponent):

\(\displaystyle 4^x=4096\)

\(\displaystyle \\4^x=16^3 \\4^x=(4^2)^3 \\4^x=4^6 \\x=6\)

Step 1: Write the expression in exponential form...

Given: \(\displaystyle \log_a b=x\implies a^x=b\)

\(\displaystyle \log_6 46656=x\implies 6^x=46656\)

Step 2: Convert the right hand side into a power of 6..

\(\displaystyle 46656=6^6\)

Step 3: Re-write the equations...

\(\displaystyle 6^x=6^6\)

Since the bases are equal, taking log of both sides will cancel them.

\(\displaystyle \log_6 6^x=\log_6 6^6\)

\(\displaystyle \implies x=6\)

\(\displaystyle In\ order\ to\ solve\ for\ x\ we\ must\ know\ how\ to\ rewrite\ a\ log\ in\ exponent\ form:\)

\(\displaystyle We\ can\ use\ the\ general\ rule:\ log_{b}(a)=x\\ when\ rewritten\ in\ exponential\ form\ we\ get: b^{x}=a\)

\(\displaystyle To\ Solve\ for\ x:\ log_{2}(8)=x;\ we\ rewrite\ as\ 2^{x}=8\)

\(\displaystyle We\ know\ that\ 2^{3}=8;\ so\ the\ value\ of\ x=3\)

Recall a few properties of logarithms:

1.When adding logarithms of like base, we multiply the inside.

2.When subtracting logarithms of like base, we divide the inside.

3. When multiplying a logarithm by a number, we can raise the inside to that power.

So we begin with this:

\(\displaystyle 5log_35x+log_3\frac{1}{4x^3}-log_3x^2\)

I would start with 3 to simplify the first log.

\(\displaystyle log_3(5x)^5+log_3\frac{1}{4x^3}-log_3x^2\)

\(\displaystyle log_33125x^5+log_3\frac{1}{4x^3}-log_3x^2\)

Next, use rule 1 on the first two logs.

\(\displaystyle log_3\left(3125x^5\cdot\frac{1}{4x^3}\right)-log_3x^2\)

\(\displaystyle log_3\left(\frac{3125x^2}{4}\right)-log_3x^2\)

Then, use rule 2 to combine these two.

\(\displaystyle log_3\left(\frac{3125x^2}{4x^2}\right)=log_3\left(\frac{3125}{4}\right)\approx 6.06\)

So our answer is 6.06.

When combining logarithms into one log, we must remember that addition and multiplication are linked and subtraction and division are linked. 

\(\displaystyle log(ab)=log(a)+log(b)\)

\(\displaystyle log\left (\frac{a}{b} \right )=log(a)+log(b)\)

In this case we have multiplication and division - so we assume anything that is negative, must be placed in the bottom of the fraction. 

\(\displaystyle [log(x)+log(z)]-log(y)=log(xz)-log(y)\)

\(\displaystyle log(xz)-log(y)=log\left ( \frac{xz}{y} \right )\)

When rewriting an exponential function as a log, we must follow the model below: 

\(\displaystyle log_{base}(result)=exponent\)

A log is used to find an exponent. The above corresponds to the exponential form below: 

\(\displaystyle base^(^e^x^p^o^n^e^n^t^)=result\)

\(\displaystyle 12^(^x^+^2^)=y\ rewritten\ in\ log\ form\ becomes:\)

\(\displaystyle log_{12}(y)=x+2\)

\(\displaystyle The\ base\ is\ 12,\ the\ exponent\ is\ (x+2)\ and\ the\ result\ is\ y.\)

In order to rewrite a log, we must remember the pattern that they follow below: 

\(\displaystyle log_{base}(result)=exponent\)

In this question we have: 

\(\displaystyle log_{3}(22)=y\)

\(\displaystyle The\ base\ is\ 3,\ the\ exponent\ is\ y\ and\ the\ result\ is\ 22.\)

\(\displaystyle We\ now\ rewrite\ this\ as:\ 3^y=22\)

Step 1: Recall all logarithm rules:

\(\displaystyle \log_a b+\log_a c=\log_a (bc)\)

\(\displaystyle \log_a b-\log_a c=log_a \bigg(\frac{b}{c}\bigg)\)

\(\displaystyle c\log_a b=\log_a (b^c)\)

\(\displaystyle \large a^{\frac {1}{b}}=\sqrt[b]{a}\)

Step 2: Rewrite the first term in the expression..

\(\displaystyle 4\log_4 x^2=\log_4 (x^2)^4=log_4 (x^8)\)

Step 3: Re-write the third term in the expression..

\(\displaystyle \frac {1}{5}\log_4 (z)=\log_4 (z)^{\frac {1}{5}}=log_4 (\sqrt[5] {z})\)

Step 4: Add up the positive terms...

\(\displaystyle \log_4(x^8)+\log_4 (\sqrt[5]{z})=\log_4 (x^8\sqrt[5]{z})\)

Step 5: Subtract the answer the other term from the answer in Step 4.

\(\displaystyle \log_4 (x^8\sqrt[5]{z})-\log_4(y^3)=\log_4 \bigg(\frac {x^8\sqrt[5]{z}}{y^3}\bigg)\)

In order to expand this log, we must remember the log rules. 

\(\displaystyle Similar\ to\ the\ exponent\ rules\ when\ variables\ are\ multiplied\ within\ a\ log:\\ the\ variables\ are\ added: log(xy)=log(x)+log(y)\)

\(\displaystyle When\ variables\ are\ divided\ within\ a\ log\ they\ are\ subtracted:\\ log\left ( \frac{x}{y} \right )=log(x)-log(y)\)

\(\displaystyle For\ this\ problem\ we\ combine\ both\ rules:\)

\(\displaystyle Expand: log\left ( \frac{xy}{z} \right )=log(x)+log(y)-log(z)\)