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Problem 1

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Problem 2:

IF 5^X = 15 then what would be 5^2x-1

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    Dear Ben, what are your ideas on problem number two? Can you use division rules to express $5^{2x-1}$ in terms of $5^x$? – 2017-02-14
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    How come $\log1$ suddenly turned into $\log10$? What is $\log1$ anyway? – 2017-02-14
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    Is it $5^{2x}-1$ or $5^{2x-1}$? – 2017-02-14
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    As well as other errors, in lines 4 and 5 it looks like you think if logarithms are too hard you can just drop them. You **seriously** need to talk to your teacher about this, you are not going to understand it from online replies. – 2017-02-14

1 Answers 1

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Problem 2:

$5^{2x-1} = \frac{(5^x)^2}{5}$

So if you know $5^x$, you can get the rest.

Problem 1:

$\log(1) = 0 $ So any real solutions come from $x^2+4x-60 = 0$ or $x^2 -5x +5 = 1$