Since the question gives,

$\displaystyle \log4=0.602$ 

and 

$\displaystyle \log12=1.079$

To evaluate 

$\displaystyle \log64$

manipulate the expression to use what is given.

$\displaystyle =\log4^3$

$\displaystyle =3\log4$

$\displaystyle =3*0.602$

$\displaystyle =1.806$

According to log rules, when an exponential is raised to the power of a logarithm, the exponential and log will cancel out, leaving only the power.

Simplify the given expression.

$\displaystyle 2e^{ln(2x+1)} = 2(2x+1)$

Distribute the integer to both terms of the binomial.

The answer is:  $\displaystyle 4x+2$

The natural log has a default base of $\displaystyle e$.

This means that the expression written can also be:

$\displaystyle 5ln_e(e^{5(5-x)})$

Recall the log property that: 

$\displaystyle ln_k(k^n) = n$

This would eliminate both the natural log and the base, leaving only the exponent.

The natural log and the base $\displaystyle e$ will be eliminated.

The expression will simplify to:

$\displaystyle 5[5(5-x)] = 5[25-5x] = 125-25x$

The answer is:  $\displaystyle 125-25x$

The log property need to solve this problem is:

$\displaystyle b^{log_b x} = x$

The base and the log of the base are similar.  They will both cancel and leave just the quantity of log based two.

$\displaystyle 2^{log_2(5x-6)}=5x-6$

The answer is:  $\displaystyle 5x-6$

Rewrite the log so that the terms are in a fraction.

$\displaystyle log_{4}(8^3) = \frac{log(8^3)}{log( 4)}$

Both terms can now be rewritten in base two.

$\displaystyle \frac{log(8^3)}{log( 4)} = \frac{log((2^3)^3)}{log( 2^2)} = \frac{log(2)^9}{log(2)^2}$

The exponents can be moved to the front as coefficients.

$\displaystyle \frac{log(2)^9}{log(2)^2} = \frac{9log(2)}{2log(2)} = \frac{9}{2}$

The answer is:  $\displaystyle \frac{9}{2}$

By the Logarithm of a Power Property, for all real $\displaystyle b$, all $\displaystyle N > 0$, $\displaystyle a > 0 , a \ne 1$

$\displaystyle b \log_{a} N = \log_{a}( N ^{b})$

Setting $\displaystyle b = \frac{1}{5} , a = 5$, the above becomes 

$\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \left ( N^{\frac{1}{5}}\right )$

Since, for any $\displaystyle a,b$ for which the expressions are defined, 

$\displaystyle N^{\frac{1}{b}} = \sqrt[b]{N}$,

setting $\displaystyle b = 5$, th equation becomes

$\displaystyle \frac{1}{5} \log _{5}N = \log_{5} \sqrt[5]{N}$.

Due to the following relationship:

$\displaystyle \sqrt[N]{7} = 7 ^{\frac{1}{N}}$; therefore, the expression 

$\displaystyle \log_{7} \sqrt[N]{7}$

can be rewritten as 

$\displaystyle \log_{7} \left ( 7 ^{\frac{1}{N}} \right )$

By definition,  

$\displaystyle \log_{a} \left (a^{M} \right )= M$.

Set $\displaystyle a = 7$ and $\displaystyle M = \frac{1}{N}$, and the equation above can be rewritten as

$\displaystyle \log_{7} \left ( 7 ^{\frac{1}{N}} \right )= \frac{1}{N}$,

or, substituting back,

$\displaystyle \log_{7} \sqrt[N]{7}= \frac{1}{N}$

To solve for $\displaystyle x$, first convert both sides to the same base:

$\displaystyle 5^{x} = \frac{1}{25}^{4x-1} = (5^{-2})^{4x-1} = 5^{-8x+2}$

Now, with the same base, the exponents can be set equal to each other:

$\displaystyle x= - 8x +2$

Solving for $\displaystyle x$ gives:

$\displaystyle x = \frac{2}{9}$

$\displaystyle \log{_{4}}{x^{2}}+\log{_{4}}{3}=1$

$\displaystyle \Rightarrow \log{_{4}}{x^{2}}+\log{_{4}}{3}=\log{_{4}}{4}$

$\displaystyle \Rightarrow log{_{4}}{(x^{2}\times 3)}=\log{_{4}}{4}$

$\displaystyle \Rightarrow 3x^{2}=4$

$\displaystyle \Rightarrow x^{2}=\frac{4}{3}$

$\displaystyle \Rightarrow x=\frac{2\sqrt{3}}{3}, x=-\frac{2\sqrt{3}}{3}$

Rewrite $\displaystyle 24$ as a product that includes the number $\displaystyle 3$:

$\displaystyle \small \log_2 24 = \log_2 (2^3 \cdot 3)$

Then we can split up the logarithm using the Product Property of Logarithms:

$\displaystyle \small \log_2 (2^3\cdot 3) = \log_2 2^3 + \log_2 3$

                     $\displaystyle \small =3+\log_2 3$

                     $\displaystyle \small \approx 3 + 1.5850$

Thus,

$\displaystyle \small \log_2 24 \approx 4.5850$.