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Hello I have the following question I have to find the answer of this logarithmic formula

$$(\log(2^5) + \log(4^{0.2}))\times(\log(5^2) + \log(25^{0.5})) .$$

I am currently having problems with the steps for the solution, and any assistance would be significantly appreciated.

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  • $\begingroup$ What don't you understand? What's the base of the logarithm here? $\endgroup$ – Dbchatto67 Apr 19 at 10:58
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I have considered base same for everything.

(5log(2)+2*0.2log(2)) x (2log(5)+2*0.5log(5))

(5.4log(2) x 3log(5))

Ans = 16.2 x log(2) x log(5)

formulas for above steps:

log ab=b log a

x log a + y log a = (x+y) log a (same base)

log (ab)c = log abc

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It is $$\frac{5\log(2)+\frac{2}{5}\log(2)}{2\log(5)+\log(5)}$$ using $$\log(x^r)=r\log(x)$$ for $x>0$

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