Logarithms are another way of writing exponents. In the general case, \(\displaystyle \small \log_a{x}=b\) really just means \(\displaystyle \small a^{b}=x\). We take the base of the logarithm (in our case, 2), raise it to whatever is on the other side of the equal sign (in our case, 4) and set that equal to what is inside the parentheses of the logarithm (in our case, x+6). Translating, we convert our original logarithm equation into \(\displaystyle \small 2^{4}=x+6\). The left side of the equation yields 16, thus \(\displaystyle \small x=10\).

To solve this equation, remember log rules

\(\displaystyle \log_{b}x=y; b^{y}=x\).

This rule can be applied here so that

\(\displaystyle 5^3=x\)

and

\(\displaystyle 125=x.\)

To solve this equation, you must first simplify the log expressions and remember log laws (\(\displaystyle \log_{b}x=y\); \(\displaystyle b^{y}=x\)). Therefore, for the first expression, \(\displaystyle 4^{x}=64\) so \(\displaystyle x=3\). For the second expression, \(\displaystyle 3^x=81\) so \(\displaystyle x=4.\) Then, substitute those numbers in so that you get \(\displaystyle 3-4=-1\).

\(\displaystyle log_{5}(x) = 2\)

Remember that a logarithm is nothing more than an exponent. The equation reads, the logarithm (exponent) with base \(\displaystyle 5\) that gives \(\displaystyle x\) is \(\displaystyle 2\). 

\(\displaystyle x = 5^2\)

\(\displaystyle x = 25\)

Since both sides are a log with the same base, their arguments must be equal.

Setting them equal:

\(\displaystyle 4x+7=11x\)

Solve for x:

\(\displaystyle 7=7x\)

\(\displaystyle 1=x\)

Set the arguments of the natural log equal and solve for \(\displaystyle x\)

\(\displaystyle 3x-2=6x\)

\(\displaystyle -2=3x\)

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

However, you cannot take the natural log of a negative number. Therefore; there is No Solution

To solve this problem, it is easiest to start by thinking about what the logarithm means:

\(\displaystyle 2^x=24\)

Now, we take the log of both sides (either base-10 or natural):

\(\displaystyle \log(2^x)=\log24\)

Next, using the properties of logarithms, we can bring the exponent x in front of the logarithm:

\(\displaystyle x\log2=\log24\)

Finally, solve for x:

\(\displaystyle x=\frac{\log24}{\log2}\)

Logarithms are inverses of exponents. 

\(\displaystyle \log_5 25\) is asking how many fives multiplied together are equal to 25.

The formula to solve logarithmic equations is as follows.

\(\displaystyle log_ab=c\Rightarrow a^c=b\)

Applying this formula to our particular problem results in the following.

\(\displaystyle \log_5 25\Rightarrow 5^x=25\) 

\(\displaystyle 5^2\) = \(\displaystyle 5*5 = 25\)

So \(\displaystyle x= 2\).

To solve, we must keep in mind that a logarithm is asking the following:

\(\displaystyle \log_ab\) means \(\displaystyle a^x=b\), and we must find x.

First, we can rewrite the sum of the base-8 logarithms as a product (this is a property of logarithms):

\(\displaystyle \log_8(4\cdot 10)=\log_840\)

Now, we use the above formula to find what this numerically equals:

\(\displaystyle 8^x=40\)

Take the natural or common logarithm of both sides, which allows us to bring the power x in front of the logarithm:

\(\displaystyle \log8^x=\log40\)

\(\displaystyle x\log8=\log40\)

\(\displaystyle x=\frac{\log40}{\log8}\)

To solve for x, we must take the logarithm of both sides (common or natural, it doesn't matter):

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

In doing this, we can now bring the exponents in front of the logarithms:

\(\displaystyle x\log4=x^2\cdot\log3\)

Now, solve for x:

\(\displaystyle \log4=x\log3\)

\(\displaystyle x=\frac{\log4}{\log3}\)