# How do you find the solution of the exponential equation 10^(1 - x) = 6^x ?

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We have to solve 10^(1 - x) = 6^x. We can use the relation log a^b = b*log a to solve this problem.

First take the original equation:

10^(1 - x) = 6^x

take the logarithm to base 10 for both the sides

log [10^(1 - x)] = log [6^x]

=> (1 - x) log 10 = x log 6

=> 1- x = x log 6

=> x ( log 6 + 1) = 1

=> x = 1 / ( log 6 + 1)

log 6 + 1 = 1.778

=> x = .5623

**Therefore the required value of x is 0.5623**

10^(1-x) = 6^x

First we will apply logarithm to both sides:

==> log (10^(1-x) = log 6^x

Now from logarithm properties, we know that:

log a^b = b*log a

==> (1-x) * log 10 = x*log 6

But we know that log 10 = 1

==> (1-x)*1 = x log 6

==> 1- x = x log 6

==> 1 = x log 6 + x

Now we will factor x:

==> 1 = x ( log 6 + 1)

Divide by log 6 + 1

==> x = 1/ (log 6 + 1)

**==> x = 0.5624 ( approx.)**

We can use logarithms to solve exponential equations.

We'll take the common logarthim both sides:

lg 10^(1 - x) = lg 6^x

We'll apply the power rule for logarithms:

(1-x) lg 10 = x lg 6

We'll recall that lg 10 = 1

We'll re-write the equation:

1 - x = x lg 6

We'll add x both sides:

x + xlg6 = 1

We'll factorize by x:

x(1 + lg 6) = 1

We'll divide by 1 + lg 6 = lg 10 + lg 6 = lg 60:

x = 1/lg 60

Rounded to four decimal places:

**x = 0.5624**

To solve 10^(1-x) = 6^x.

We take logarithms of both sides:

(1-x)log10 = xlog6

1-x= xlog6, as log10 = 1.

We add x to both sides:

1= x+xlog6

1= x(1+log6).

1/(1+log6) = x.

x = 1/(1+log6) = 0.5624 approximately.

Tally:

10^(1-x) = 10^(1-0.5624) .= 10^0.4376 = 2.739 approximately.

6^x = 6^0.5624 = 2.739 approximately