# Find the solution of the exponential equation 2^(1-x)=19 in terms of logarithms or correct the four decimal places.

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### 4 Answers

2^(1-x) = 19

==> To solve the equality, we will us the logarithm form.

==> First. let us apply the logarithm to both sides:

==> log 2^(1-x) = log 19

Now we know from logarithm properties that:

log a^b = b *log a

==> log 2^(1-x) = log 19

==> (1-x)*log 2 = log 19

Now we will divide by log 2

==> (1-x) = log 19/ log 2

==> -x = log 19/ log 2 - 1

Now multiply by -1.

==> x = 1 - log19/log2

==> x = 1 - 4,2479

= -3,2479

**Then, the answer is x = -3,2479.**

We need to solve 2^(1-x)=19 for x.

Now we have the equation 2^(1-x)=19

Take the logarithm of both the sides

=> log [2^(1-x)] = log 19

Now we know that log a^b = b log a

=> ( 1-x) log 2 = log 19

=> 1-x = log 19 / log 2

Now log 19 = 1.2787 and log 2 = 0.3010

=> 1 - x = 1.2787 / 0.3010

=> 1- x = 4.2479

=> x = - 3.2479

=> x = -3.2479

**Therefore the required result is -3.2479.**

To find x from 2^(1-x) = 19.

2^(1-x) = 2*2^-x = 2/2^x.

Therefore 2/2^x = 19.

2 = 19*2^x.

2^x = 2/19.

We take logarithms of both sides:

x log2 = log(2/19) = log2 - log19.

x = {log2 - log19)/log2 = 1- log19/log2.

x = 1- 4.24793 = -3.24793

x= -3.2479 for 4 decimal places.

x =

x log2 =

We'll take log on both sides:

log 2^(1-x)=log 19

We'll use the power rule of logarithms:

(1-x)*log 2 = log 19

We'll divide by log2 both sides of the equation:

(1-x) = log 19 / log 2

We'll subtract 1 both sides:

-x = log 19 / log 2 - 1

We'll multiply by -1:

x = 1 - log 19 / log 2

We'll compute the ratio log 19 / log 2 = 4.2479

x = 1 - 4.2479

**x = -3.2479**