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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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.
Bill Dubuque
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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
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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
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Recall that $\rm\ a^{log_a x}\ =\ x\ $ from your post a few hours ago. – Bill Dubuque Nov 18 '10 at 16:21
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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
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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
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Hint: 3y + y = 64
Further (basic) hint: $a^{\log _b c} = (e^{\log a} )^{\log c/\log b} $
Shai Covo
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N[3*(5^4)^(Log [5, 2]) + 2^(Log[5, 5^4]) - 64]gives 0**. – Quixotic Nov 18 '10 at 16:08