Question #9ce73

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Question #9ce73

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Answer 1 · 2017-07-01T08:48:11

This equality cannot be solved for $t$ $t$ as there are no real solutions for $t$ $t$ . However, there are possible complex solutions.

Explanation:

Using log laws we can simplify somewhat and get the $t$ $t$ 's on the same side:

${log}_{5} (3 \cdot 7^{t}) = {log}_{5} (5^{2^{t}})$ $log_5(3*7^t)=log_5(5^(2^t))$

$\Rightarrow {log}_{5} (3 \cdot 7^{t}) = 2^{t}$ $rArrlog_5(3*7^t)=2^t$

$\Rightarrow {log}_{5} (3) + {log}_{5} (7^{t}) = 2^{t}$ $rArrlog_5(3)+log_5(7^t)=2^t$

$\Rightarrow {log}_{5} (3) + t \cdot {log}_{5} (7) = 2^{t}$ $rArrlog_5(3)+t*log_5(7)=2^t$

$\Rightarrow {log}_{5} (3) = 2^{t} - t \cdot {log}_{5} (7)$ $rArrlog_5(3)=2^t-t*log_5(7)$

I don't even think this form is simpler than the original, I'm not sure if it's worth doing... But more importantly, how do we know whether we can actually solve for $t$ $t$ ?

Let's set:

$y_{1} = 3 \cdot 7^{t}$ $y_1=3*7^t$

$y_{2} = 5^{2^{t}}$ $y_2=5^(2^t)$

Setting the two equations equal to each other will allow you to solve for the t values where the two graphs intersect (this is what the original question asked). You should be able to visually find these as the point(s) where the two graphs cross each other.

$y_{1} = y_{2}$ $y_1=y_2$

$\Rightarrow 3 \cdot 7^{t} = 5^{2^{t}}$ $rArr3*7^t=5^(2^t)$

So let's plot the two together and have a look.

As t goes to negative $\infty$ $oo$ , $y_{1}$ $y_1$ approaches zero and $y_{2}$ $y_2$ approaches 1. They never cross. As t goes to positive $\infty$ $oo$ the two graphs diverge once again and never cross. Because these two graphs never intersect, there aren't any real solutions for t.

So, I can't simplify the original equality any further and solve for $t$ $t$ , unless we go into complex land. But I dare not tread there!

Answer 2 · 2017-07-01T10:31:10

See below.

Explanation:

Applying $log$ $log$ to both sides

$log 3 + log 7 \cdot t = 2^{t} log 5$ $log3+log7*t=2^t log5$

This equation has the structure

$a t + b = 2^{t}$ $at+b=2^t$

with $a = \frac{log 7}{log 5}$ $a=(log7)/(log5)$ and $b = \frac{log 3}{log 5}$ $b = log3/log5$

Now using the Lambert function

https://en.wikipedia.org/wiki/Lambert_W_function

such that

$y = x e^{x} \Leftrightarrow x = W (y)$ $y=xe^x hArr x=W(y)$

We will transform

$a t + b = 2^{t}$ $at+b=2^t$

into a suitable form to be handled with the Lambert function so making $ξ = - (t + \frac{b}{a})$ $xi = -(t + b/a)$

$a t + b = 2^{t} \Rightarrow ξ 2^{ξ} = r = - \frac{1}{a} 2^{- \frac{b}{a}} \Rightarrow ξ = \frac{W (r log 2)}{log 2}$ $at+b=2^t rArr xi 2^xi = r = -1/a2^(-b/a) rArr xi=(W(r log2))/log2$

then

$t = - \frac{W (- \frac{log 2}{a} 2^{- \frac{b}{a}})}{log 2} - \frac{b}{a}$ $t = - (W(-log2/a 2^(-b/a)))/log2-b/a$ or

$t = - \frac{log 3}{log 7} - \frac{W (- \frac{2^{- (\frac{log 3}{log 7})} log 2 log 5}{log 7})}{log 2}$ $t=- Log3/log 7- (W(-(2^(-(Log3/Log7)) Log2 Log5)/Log7))/log2$

This gives a complex solution which is

$t = 0.827751 - 0.465279 i$ $t=0.827751 - 0.465279 i$

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Question #9ce73

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