\(\displaystyle 2^x=128\)

\(\displaystyle 2^x=2^7\)

\(\displaystyle \log_22^x=\log_22^7\)

\(\displaystyle x=7\)

We notice that all of the bases are powers of 2 so we can re-write them as 2 raised to a certain power. 

\(\displaystyle 2^x(2^2)^2=(2^3)^{3+x}\)

Then we simplify both sides of the equation so we have a single base 2 raised to a power. When we have when we have exponentials with the same base multiplied we add the exponents, when we have them raised to an exponent we multiply the exponents. 

\(\displaystyle 2^{x+4}=2^{9+3x}\)

Now that we have the same base on both sides we set the exponents equal and solve for x.

\(\displaystyle x+4=9+3x\)

\(\displaystyle 2x=-5\)

\(\displaystyle x=\frac{-5}{2}\)

The expression can be simplified using logarithmic properties.

\(\displaystyle \log(x^2+10x+25) +\log (3x+15)=4 + \log 3 \\\)

\(\displaystyle \log(x+5)^2 +\log (3*(x+5))=4 + \log 3\)

\(\displaystyle 2\log(x+5) +\log 3 + \log(x+5)= 4 + \log 3\)

\(\displaystyle 3 \log(x+5)=4\)

We use the definition of common logarithm to solve for x.

\(\displaystyle (x+5)^3=10^4\)

\(\displaystyle (x+5)=\sqrt[3]{10000}=21.544\)

So x = 21.544 -5 or 16.544.

Log Property: Log(x) + Log(y) = Log(xy)

The equation \(\displaystyle log (x)+log(x+3) =1\) can be simplified using the log property stated above. 

The simplified version of the equation is:

\(\displaystyle log((x)(x-3))=1\)

Substitute Log(10) in for 1, since Log(10) = 1

\(\displaystyle log((x)(x-3))=log(10)\)

Next, remove the logs from both sides

\(\displaystyle (x)(x-3)=10\)

Solve for the value of x:

\(\displaystyle x(x-3)=10\)

\(\displaystyle x^{2}-3x=10\)

\(\displaystyle x^{2}-3x-10=0\)

\(\displaystyle (x-5)(x+2)=0\)

\(\displaystyle x=5, x=-2\)

A negative number cannot be evaluated in the log function equation above so the answer is \(\displaystyle x=5\)

First, remove the log from the equation by changing the form:

\(\displaystyle log{_{16}}{4^{x}}=2\)

\(\displaystyle 16^{2}=4^{x}\)

Next, change both sides so that they share a common base. In this case the common base is 4.

\(\displaystyle (4^{2})^{2}=4^{x}\)

Reduce the amount of exponents on the left side. Since an exponent is raised to an exponent you multiply the exponents together.

\(\displaystyle 4^{4}=4^{x}\)

Set the exponents on both sides equal to eachother and solve for x.

\(\displaystyle 4=x\)

Answer: \(\displaystyle x=4\)

Since the logarithmic function \(\displaystyle y=\log_{b}(x)\) is the inverse of the exponential function \(\displaystyle x=b^y\), we can make a straightforward simplification as follows:

\(\displaystyle \frac{1}{64}=16^x\)

\(\displaystyle 64^-^1=16^x\)

\(\displaystyle 2^-^6=2^{4x}\)

\(\displaystyle -6 \log2=4x \log2\)

\(\displaystyle -6=4x\)

\(\displaystyle x=-\frac{3}{2}\)

Using the logarithmic product rule \(\displaystyle \log_b(xy)= \log_bx + \log_b y\), we simplify as follows:

\(\displaystyle \log_{8}2+\log_{8}4+\log_{8}8 = \log_8(8\cdot4\cdot2)\)

\(\displaystyle \log_864\)

\(\displaystyle 8^x=64\)

\(\displaystyle x=2\)

Utilizing the power rule of logs we reduce the original expression to:

 \(\displaystyle \frac{1}{2}{}\log_4{16x}\)

From there we use the product property to obtain:

 \(\displaystyle \frac{1}{2}(\log_416+log_4 x)\)

Since 4 to the second power is 16 this expression yields 

\(\displaystyle \frac{1}{2}(2+log_4 x)\) , then we multiply through by the one half to obtain our final answer \(\displaystyle \left(1+\frac{1}{2}log_4 x\right)\).

First use the power property of logs to get:

 \(\displaystyle e log_a\frac{b}{cd}\).

Then use the quotient property to get:

 \(\displaystyle e (log_a b -log_acd)\).

Then finally use the product property and simplify to yield the final answer, \(\displaystyle e (log_a b -log_ac-log_ad)\).

\(\displaystyle \frac{8^49^3}{2^{10}3^4}\) 

We observe that \(\displaystyle 8=2^3\) and \(\displaystyle 9=3^2\) and substitute to yeild

\(\displaystyle \frac{(2^3)^4(3^2)^3}{2^{10}3^4}\). 

Now we use the fact that \(\displaystyle (a^b)^c=a^{b*c}\) to simplify  

\(\displaystyle \frac{2^{12}3^6}{2^{10}3^4}\). 

Since \(\displaystyle \frac{a^b}{a^c}=a^{b-c}\) simplifying again yeilds

\(\displaystyle 2^23^2\) which is just \(\displaystyle 4\cdot9\)

This yeilds a final answer of 36