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 in exponential form:

\(\displaystyle 9^{2}=2x+1\)

Solve for x:

\(\displaystyle 81=2x+1\)

\(\displaystyle x=40\)

For this problem it is helpful to remember that,

 \(\displaystyle ln(3x+8) = 0\) is equivalent to \(\displaystyle ln(3x+8) = ln(1)\) because \(\displaystyle ln(1)=0\)

Therefore we can set what is inside of the parentheses equal to each other and solve for \(\displaystyle x\) as follows:

\(\displaystyle 3x+8=1\)

\(\displaystyle 3x=-7\)

\(\displaystyle x=-\frac{7}{3}\)

To solve this problem you must be familiar with the one-to-one logarithmic property.

\(\displaystyle \log_{10}x=\log_{10}y\) if and only if x=y. This allows us to eliminate to logarithmic functions assuming they have the same base.

\(\displaystyle \log_{10}(5x-2)=\log_{10}(x+6)\)

one-to-one property:

\(\displaystyle (5x-2)=(x+6)\)

isolate x's to one side:

\(\displaystyle 4x-2=6\)

move constant:

\(\displaystyle 4x=8\)

\(\displaystyle x=2\)

Get all the terms with e on one side of the equation and constants on the other.

\(\displaystyle 5e^{3x+4}-10=0\)

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

Apply the logarithmic function to both sides of the equation.

\(\displaystyle ln(e^{3x+4})=ln2\)

\(\displaystyle 3x+4=ln2\)

\(\displaystyle 3x=-4+ln2\)

\(\displaystyle x=\frac{-4+ln2}{3}\)

Recall the rules of logs to solve this problem. 

First, when there is a coefficient in front of log, this is the same as log with the inside term raised to the outside coefficient.

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

\(\displaystyle log(x^{2})+log(5)=3\)

Also, when logs of the same base are added together, that is the same as the two inside terms multiplied together.

In mathematical terms:

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

Thus our equation becomes,

\(\displaystyle log(5x^{2})=3\)

To simplify further use the rule, 

\(\displaystyle log_b(x)\rightarrow b^{log_b(x)}=x\)

\(\displaystyle 10^{log(5x^{2})}=10^{3}\)

\(\displaystyle 5x^{2}=1000\)

\(\displaystyle x^{2}=200\)

\(\displaystyle x=\pm10\sqrt{2}\).

Solve the following equation for t

\(\displaystyle log_t(625)=4\)

We can solve this equation by rewriting it as an exponential equation:

\(\displaystyle t^4=625\)

Next, take the fourth root to get:

\(\displaystyle t=\sqrt[4]{625}=5\)

We can check our work vie the following:

\(\displaystyle 5^4=625\)

\(\displaystyle \ln(x^2+1)=3\)

The first step step is to carry out the inverse operation of the natural logarithm, 

\(\displaystyle e^{\ln(x^2+1)}=e^3\)

We can use the property \(\displaystyle e^{ln(u)}=u\) to simplify the left side of the equation to obtain,  

\(\displaystyle x^2+1 = e^3\)

Solve for \(\displaystyle x\), 

\(\displaystyle x = \pm \sqrt{e^3-1}\)

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

Note that this equation is a quadratic because \(\displaystyle e^{2x}=(e^x)^2\)

Factor:

\(\displaystyle (e^x-2)(e^x-1)\)

Set each factor equal to 0 and solve:

\(\displaystyle e^x-2=0\rightarrow e^x=2 \rightarrow \boldsymbol{x=ln\hspace{1mm} 2}\)

\(\displaystyle e^x-1=0\rightarrow e^x=1 \rightarrow \boldsymbol{x=0}\)