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Solve the equation for $x$ by using base 10 logarithms.

$$16\cdot4^{2.5x}=9$$
EDIT: I made a typo (somehow... I was very far off!!)
The correct equation is this: $$16\cdot4^{2.5x}=70$$

Can it be written like:
$$2.5x\log_{10}(5)=70\ ?$$

Then get:
$$\log_{10}(5)=\frac{70}{2.5x}$$
The computer wants a largest value and smallest value, similar to an answer for a quadratic problem. I need to know how to get the 2 answers even if one ends up negative (I know the negative will be tossed out, but I still need to know how to get the answer).

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  • $\begingroup$ $16\cdot 4^k=4^{k+2}=2^{2k+4}$... $\endgroup$ – abiessu May 7 '14 at 2:53
  • $\begingroup$ I don't think that helps my case with logarithms. $\endgroup$ – GeekyDewd May 7 '14 at 2:55
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$$ \begin{align} 16\cdot4^{2.5x}&=70\\ 2^4\cdot(2^2)^{2.5x}&=70\\ 2^4\cdot2^{5x}&=70\\ 2^{4+5x}&=70\\ \log_{10}2^{4+5x}&=\log_{10}70\\ (4+5x)\log_{10}2&=\log_{10}70\\ 4+5x&=\frac{\log_{10}70}{\log_{10}2} \end{align} $$ Can you take it from here?

Addendum : $$ \begin{align} 4^2\cdot(2^2)^{2.5x}&=70\\ 4^2\cdot(2^2)^{2.5x}-(\sqrt{70})^2&=0\\ (4\cdot2^{2.5x}-\sqrt{70})(4\cdot2^{2.5x}+\sqrt{70})&=0 \end{align} $$ It will yield two solutions like you want.

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  • $\begingroup$ I'm not sure the sum side of the quadratic will produce a unique result for $x$. Is it possible that what is being asked for are upper and lower estimates of the decimal value of $x$? $\endgroup$ – abiessu May 7 '14 at 12:34
  • $\begingroup$ @abiessu I know that it's weird but the question is also weird since the OP asks for "get the 2 answers even if one ends up negative (I know the negative will be tossed out, but I still need to know how to get the answer)". $\endgroup$ – Tunk-Fey May 7 '14 at 12:37
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    $\begingroup$ I'd like to know how you would compute the second "solution", as that would need the logarithm of a negative number. Just out of curiosity... Of course there is only one (real) solution. $\endgroup$ – Jean-Claude Arbaut May 7 '14 at 12:56
  • $\begingroup$ @Jean-ClaudeArbaut Please read my comment above your comment & also the OP. $\endgroup$ – Tunk-Fey May 8 '14 at 2:59
  • $\begingroup$ @Jean-ClaudeArbaut I later came to realize that only one answer is needed for this problem. But for problems with a radical then I would need positive and negative. I apologize for the confusion, but I didn't understand anything about the problem until I got some help. $\endgroup$ – GeekyDewd May 8 '14 at 4:36
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$16 \cdot 4^{2.5x} = 16 \cdot (4^{2.5})^x = 16 \cdot (4^{\frac{5}{2}})^x = 16 \cdot 32^x$.

So, $32^x = \frac{9}{16}$.

Thus, $x = \log_{32}(\frac{9}{16}) = \dfrac{\log_{10}(\frac{9}{16})}{\log_{10}(32)}$

For the updated equation

$16 \cdot 4^{2.5x} = 16 \cdot (4^{2.5})^x = 16 \cdot (4^{\frac{5}{2}})^x = 16 \cdot 32^x$.

So, $32^x = \frac{30}{16}$.

Thus, $x = \log_{32}(\frac{30}{16}) = \dfrac{\log_{10}(\frac{15}{8})}{\log_{10}(32)}$

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  • $\begingroup$ I appreciate your answer, I made an edit to the original post. According to the computer that is not complete, I need two answers (one may be negative and not be used) they seem to be in decimal form. $\endgroup$ – GeekyDewd May 7 '14 at 3:27
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16*4^(2.5*x) = 70 can be written as 4^(2)*4^(2.5*x) = 70

i.e. 4^(2+2.5x) = 70

Taking log base 10 both sides

=> log(4^(2+2.5*x)) = log(70)

=> (2+2.5*x)*log(4) = log(70)

=> 2+2.5*x = log(70)/log(4)

=> 2.5*x = log(70)/log(4) -2

=> x = (log(70)/log(4) - 2) / 2.5

i.e (log(70) base 4) -2 / 2.5

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You can solve for x by using this property of logarithms.

$$log_b(x^{n}) = nlog_b(x)$$

Work:

$$16*4^{2.5x} = 70$$

$$4^{2.5x} = 70/16$$

$$log(4^{2.5x}) = log(70/16)$$

$$(2.5x)log(4) = log(70/16)$$

$$x = log(70/16)/(2.5log(4) = 0.425857...$$

Thus

$$16 * 4^{2.5*0.425857...} = 70$$

More info here: http://www.andrews.edu/~calkins/math/webtexts/numb17.htm

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Taking the updated equation,

$$16\cdot 4^{2.5x}=70\\ 2^{5x}=\frac{35}8\\ 5x\log_{10} 2=\log_{10} 7+\log_{10} 5-\log_{10} 8\\ x={\log_{10} 7+1-4\log_{10} 2\over5\log_{10} 2}\\ x={\log_2 7+\log_2 5-3\over 5}$$

It should be fairly apparent how to rewrite in terms of other logarithm bases.

One possible interpretation of the smallest and largest values is as upper and lower estimates of the decimal value of $x$.

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  • $\begingroup$ Not sure how you pass from the third line to the fourth. $\endgroup$ – Jean-Claude Arbaut May 7 '14 at 12:57
  • $\begingroup$ Ah, I missed an extra log factor, thanks. $\endgroup$ – abiessu May 7 '14 at 15:03

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