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

Question

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).

Answer 1

$$ \begin{align} 16\cdot4^{2.5x}&=70\\ 2^4\cdot(2^2)^{2.5x}&=70\\ 2^4\cdot2^{5x}&=70\\ 2^{4+5x}&=70\\ \log_{10}2^{4+5x}&=\log_{10}70\\ (4+5x)\log_{10}2&=\log_{10}70\\ 4+5x&=\frac{\log_{10}70}{\log_{10}2} \end{align} $$ 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.

Answer 2

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

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

Answer 3

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 $x$.

Answer 4

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

Answer 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

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

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