We begin with the following:

$\displaystyle 343=7^{15t-12}$

This can be rewritten as

$\displaystyle 7^3=7^{15t-12}$

Recall that if you have two exponents with equal bases, you can simply set the exponents equal to eachother. Do so to get the following:

$\displaystyle 3=15t-12$

Solve this to get t.

$\displaystyle 15=15t$

$\displaystyle t=1$

We need to make the bases equal before attempting to solve for $\displaystyle x$. Since $\displaystyle 16=4^2$ we can rewrite our equation as

$\displaystyle 4^{2x}=4^{2^{2x-6}}$

    Remember: the exponent rule $\displaystyle (x^n)^m=x^{n\cdot m}$

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

Now that our bases are equal, we can set the exponents equal to each other and solve for $\displaystyle x$. 

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

$\displaystyle 2x=4x-12$

$\displaystyle 12=2x$

$\displaystyle x=6$

The first step is to make sure we don't have a zero on one side which we can easily take care of: 

$\displaystyle 7^{2y}=3^{y+2}$

Now we can take the logarithm of both sides using natural log:

$\displaystyle \ln 7^{2y}=\ln 3^{y+2}$

Note: we can apply the Power Rule here $\displaystyle \ln (x^y)=y\ln (x)$

$\displaystyle 2y \ln 7=(y+2)\ln3$

$\displaystyle 2y \ln 7=y\ln3+2\ln3$

$\displaystyle 2y \ln 7-y\ln3=2\ln3$

$\displaystyle y(2 \ln 7-\ln3)=2\ln3$

$\displaystyle y=\frac{2\ln3}{2 \ln 7-\ln3}$

Before beginning to solve for $\displaystyle x$, we need $\displaystyle e$ to have a coefficient of $\displaystyle 1$: 

$\displaystyle e^{(4x-5)}=8$

Now we can take the natural log of both sides:

$\displaystyle \ln e^{(4x-5)}=\ln8$

Note: $\displaystyle \ln (e)=1$

$\displaystyle 4x-5=\ln8$

$\displaystyle 4x=\ln8+5$

$\displaystyle x=\frac{\ln8+5}{4}$

$\displaystyle 6^{3} \times 6^{n} =6^{12}$

Since the base is $\displaystyle 6$ for both, then:

$\displaystyle 3 + n =12$  When the base is the same, and you are multiplying, the exponents are added.

$\displaystyle n= 9$

$\displaystyle 10^{3x} = 63$

To solve, use common $\displaystyle \log$

$\displaystyle \log (10^{3x}) = \log (63)$

$\displaystyle 3x\log (10) = \log (63)$

$\displaystyle 3x (1) = \log (63)$

$\displaystyle 3x = \log (63)$

$\displaystyle x = \frac{\log (63)}{3}$

$\displaystyle 2^{x} = 40$

To solve, use the natural log.

$\displaystyle ln (2^{x}) = ln (40)$

$\displaystyle xln (2) = ln(40)$

To isolate the variable, divide both sides by $\displaystyle ln (2)$.

$\displaystyle x = \frac{\\ln (40)}{\\ln (2)}$

$\displaystyle 4^{x} = 200$

To solve, use the natural log.

$\displaystyle \\ln (4^{x}) = \ln(200)$

$\displaystyle x\ln (4) = \ln (200)$

$\displaystyle x = \ln (200)$

$\displaystyle x = \frac{ln(200)}{ln (4)}$

$\displaystyle -2 (10^{3x}) = -500$

Isolate the exponential part of the equation.

$\displaystyle \frac{-2 (10^{3x})}{-2} = \frac{-500}{-2}$

$\displaystyle 10^{3x} = 250$

Convert to log form and solve.

$\displaystyle \log_{10} (250) = 3x$

$\displaystyle \log_{10} (250)$ can also be written as $\displaystyle \log 250$.

$\displaystyle \frac{\log (250) }{3} = x$ 

$\displaystyle 8\times e^{0.3x} = 12$

Simply the exponential part of the equation by dividing both sides by$\displaystyle 8.$

$\displaystyle e^{0.3x} = \frac{12}{8}$

$\displaystyle e^{0.3x} = 1.5$

Write in logarithm form.

$\displaystyle \log_{e} (1.5) = 0.3x$

Because $\displaystyle \log_{e} (1.5)$ is also written as $\displaystyle \\ln 1.5$

$\displaystyle \frac{\\ln 1.5}{0.3} = x$