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I am facing some trouble in solving this equation:

$$ 3\cdot x^{\log_5 2} + 2^{\log_5 x} = 64 $$

Give me some hints to proceed on this.

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@JM: That's as copied from my module,I am not getting the explanation since it is taking $x = 5^{\log_5 x}$ they somehow reduced it to $5^4$ which I just now verified In mathmatica N[3*(5^4)^(Log [5, 2]) + 2^(Log[5, 5^4]) - 64] gives 0**. –  Quixotic Nov 18 '10 at 16:08
    
Bill showed what you're supposed to do. As for me, it's late, I better retire for now. :) –  J. M. Nov 18 '10 at 16:12

2 Answers 2

up vote 6 down vote accepted

HINT $\rm\ \ x^{\: log_5 2}\ =\ 5^{\: log_5 x\ log_5 2}\ =\ 2^{\: log_5 x}\ \ $ (or take $\rm\:log_5\:$ of both sides if that's clearer to you)

Hence the equation reduces to $\rm\ 2^{\: \log_5 x}\ =\ 2^4\ \Rightarrow\ log_5\: x\ =\ 4\ \Rightarrow\ x\ =\ \ldots$

TIP $\ $ Typically such exponential equations will be solvable in closed form only if the exponentials are all linearly dependent, so you should always check for that first.

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Sorry, I am not getting it, could you please give a bit of more hint ? –  Quixotic Nov 18 '10 at 16:09
    
What don't you get? By my hint, the equation reduces to $\ 2^{log_5 x} = 16$ –  Bill Dubuque Nov 18 '10 at 16:14
    
Recall that $\rm\ a^{log_a x}\ =\ x\ $ from your post a few hours ago. –  Bill Dubuque Nov 18 '10 at 16:21
    
Nopes,the later part is not a problem,what I am not getting is How $a^{\log_b x} = x^{\log_b a}$ ?.So my problem is here: $\rm\ \ x^{\: log_5 2}\ =\ 5^{\: log_5 x\ log_5 2}\ =\ 2^{\: log_5 x}\ \$ –  Quixotic Nov 18 '10 at 18:29
    
The first way I mentioned uses the basic identity from your prior post. But I think the second way may be clearer for you. Simply note that both terms are equal since they have equal $\rm\: log_b's$, viz. $\rm\ log_b(a^{log_b(x)}) = log_b(a)\ log_b(x) = log_b(x^{log_b(a)}) $ –  Bill Dubuque Nov 18 '10 at 19:19

Hint: 3y + y = 64

Further (basic) hint: $a^{\log _b c} = (e^{\log a} )^{\log c/\log b} $

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I realize that $e$ shouldn't be used in our context. –  Shai Covo Nov 18 '10 at 16:35

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