I don't mean to take away from anything that you've said, but in my math class, we haven't discussed what "ln" means or what it does. So, using it would probably just confuse me more, I think.

Thank you though.

Edit: Just saw that you were an Iron Maiden fan in your profile. Good call on that!

ln is a special name for the logarithm with base e.

\(\displaystyle \ln(x) = \log_e(x)\)

There is no reason why you can't use any other base such as base 10

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Edit: Same working as above but using base 10 (find and replace <3):

Example: \(\displaystyle 2^{3x-1} = 11^{x+2}\)

Use the log power law: \(\displaystyle (3x-1)\log_{10}(2) = (x+2)\log_{10}(11)\)

Distribute \(\displaystyle 3\log_{10}(2)x - \log_{10}(2) = x\log_{10}(11) + 2\log_{10}(11)\)

Simplify \(\displaystyle 3\log_{10}(2)x - x\log_{10}(11) = 2\log_{10}(11) + \log_{10}(2)\)

Factorise \(\displaystyle x(3\log_{10}(2)-\log_{10}(11)) = 2\log_{10}(11) + \log_{10}(2)\)

Solve \(\displaystyle x = \frac{2\log_{10}(11) + \log_{10}(2)}{3\log_{10}(2)-\log_{10}(11)}\)

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b) 4 \cdot 4^{x} = 5

Since the base is the same there are two ways to do this:

1)Laws of exponents say that \(\displaystyle 4 \cdot 4^x = 4^{x+1}\)

Then the same as question a)

2) \(\displaystyle 4^x = \frac{5}{4}\)

take logs: \(\displaystyle x\log_{10}(4) = \log_{10}(5)-\log_{10}(4)\)

\(\displaystyle x = \frac{\log_{10}(5)-\log_{10}(4)}{\log_{10}(4)}\)