# solve the equation using logarithms (I think this is easy level)

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).

• $16\cdot 4^k=4^{k+2}=2^{2k+4}$... – abiessu May 7 '14 at 2:53
• I don't think that helps my case with logarithms. – GeekyDewd May 7 '14 at 2:55

\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.

• 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 $x$? – abiessu May 7 '14 at 12:34
• @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)". – Tunk-Fey May 7 '14 at 12:37
• 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. – Jean-Claude Arbaut May 7 '14 at 12:56
• @Jean-ClaudeArbaut Please read my comment above your comment & also the OP. – Tunk-Fey May 8 '14 at 2:59
• @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. – GeekyDewd May 8 '14 at 4:36

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

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

• 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. – GeekyDewd May 7 '14 at 3:27

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

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

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

$$16\cdot 4^{2.5x}=70\\ 2^{5x}=\frac{35}8\\ 5x\log_{10} 2=\log_{10} 7+\log_{10} 5-\log_{10} 8\\ x={\log_{10} 7+1-4\log_{10} 2\over5\log_{10} 2}\\ x={\log_2 7+\log_2 5-3\over 5}$$
One possible interpretation of the smallest and largest values is as upper and lower estimates of the decimal value of $x$.