Solving exponentials using logs exam question help

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Jsimons56
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Attachment 774266 can someone explain how to work out part c please?
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Prince Philip
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(Original post by Jsimons56)
Attachment 774266 can someone explain how to work out part c please?
What have you tried so far?
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ManLike007
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(Original post by Jsimons56)
can someone explain how to work out part c please?
Equate the two equations, standard procedure.

6(5)^{x}=9^{x}

Apply the logs,

\log{(6(5)^{x})} = \log{(9)^{x}}

Quick note that 6(5)^x \equiv 6 \times 5^{x} so we must simplify the log,

Use these two rules,

\log(ab) \equiv \log(a) + \log(b)

\log{(a)^{b}} \equiv b \log{(a)}

Therefore,

\displaystyle \log{(6)} + \underbrace{\log{(5)^{x}}}_{x \log{(5)}} = \underbrace{\log{(9)^{x}}}_{x \log{(9)}}

Now I hope you know how to make x the subject given that I've written a note underneath the log(5) and log(9).

Now, you're not finished here.

Use this rule below last after using the two rules above first to get the required answer.

\displaystyle \log_{a}(b) \equiv \frac{\log{(b)}}{\log{(a)}}

(Alternatively, you can also have \displaystyle \log_{a}(b) \equiv \frac{\log_{x}{(b)}}{\log_{x}{(a  )}} )

Since this is maths, there's probably a faster method than mine but this was how I worked it out.
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