Question

if 2^x- 6.2 ^x+ 8 = 0 then X is​

Answer 1

Given:

The equation:

$2^x- 6.2 ^x+ 8 = 0$

To find:

$x =\ ?$

Solution:

First of all, let $2^x=t$ and let us solve for $t$, after that we will find the value of $x$.

$t -6t+8 = 0\\\Rightarrow t = \frac{8}{5}$

Now, as per our assumption:

$t=2^x=\dfrac{8}{5}$

Now, let us have a look at a few logarithmic formula as follows:

Formula of log used:

1. $log_aa=1$

2. $log(\frac{a}{b}) = loga-logb$

3. $loga^b = bloga$

Taking log base 2 both sides:

$log_22^x=log_2\dfrac{8}{5}\\\Rightarrow xlog_22=log_2\dfrac{8}-log_2{5}\\\Rightarrow x\times 1=log_2{2^3}-log_2{5}\\\Rightarrow x=3log_2{2}-log_2{5}\\\Rightarrow \bold{x=3-log_2{5}}$

Therefore the answer is:

$\bold{x=3-log_2{5}}$

Answer 2

Answer:

if 2* - 8 = 0 than x is equal to :