```
2 = \log x
```

To solve this, first exponent both sides on 10, which will reverse the log. The hard part seems to be here.```
10^2 = 10^{\log x}
```

and simplify (`10^{\log x} = x`

).```
100 = x
```

To make is harder, multiply the x by a constant inside the log.

```
2 = \log 2x
```

put both sides as exponents of 10```
10^2 = 10^{\log 2x}
```

simplify```
100 = 2x
```

divide both sides by 2```
\frac{100}{2} = \frac{2x}{2}
```

simplify```
50 = x
```

and this time add something to the x inside the log

```
2 = \log\left( 2x + 1 \right)
```

put both as exponents of 10```
10^2 = 10^{\log\left( 2x + 1 \right)}
```

simplify```
100 = 2x + 1
```

subtract 1 from both sides```
100 - 1 = 2x + 1 - 1
```

simplify```
99 = 2x
```

divide both sides by two```
\frac{99}{2} = \frac{2x}{2}
```

simplify```
\frac{99}{2} = x
```

what can I do to make it harder? Two layers of log? two layers of logs of different bases? that should be plenty hard.

```
2 = \ln\log x
```

The `\ln x`

is the outside log, so it needs to be dealt with first, make both sides exponents of `e`

```
e^2 = e^{\ln\log x}
```

simplify```
e^2 = \log x
```

make both sides exponents of 10```
10^{e^2} = 10^{\log x}
```

simplify```
10^{e^2} = x
```

And now for one that looks a bit scary, the log is in an exponent.

```
2 = e^{\log x}
```

First there is the `e^x`

to deal with, so take the natural logarithm of both sides.```
\ln 2 = \ln e^{\log x}
```

simplify```
\ln 2 = \log x
```

and put both sides on an exponent of 10, as in the previous solutions```
10^{\ln 2} = 10^{\log x}
```

simplify```
10^{\ln 2} = x
```

It is all a matter of removing one layer at a time to get to the thing being solved for.

all good?

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