How do you solve #(1) ^x = 3^(x-6)#?

Answer 1

Benjamin Achtenberg

Real solution:

#x=6#

Complex solutions:

#x = 6 + (2k pi i)/ln 3# for any integer #k in ZZ#

For any value of #x#, we have #1^x = 1#

So our equation simplifies to:

#3^(x-6) = 1#

If #x = 6# then:

#3^(x-6) = 3^0 = 1#

satisfying the equation.

As a Real valued function of Reals #f(t) = 3^t# is one to one, so this is the unique Real solution.

Complex solutions

If #k# is any integer then:

#1 = e^(2k pi i) = (e^(ln 3))^((2kpi i)/(ln 3)) = 3^((2kpi i)/(ln 3))#

So:

#3^(t+(2k pi i)/ln 3) = 3^t * 3^((2k pi i)/ln 3) = 3^t#

Hence the equation #(1)^x = 3^(x-6)# has solutions:

#x = 6 + (2k pi i)/ln 3#

for any integer #k in ZZ#

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Answer 2

Serenity Jones

To solve $(1)^x = 3^{x-6}$ , take the natural logarithm of both sides to eliminate the exponents:

$\ln(1^x) = \ln(3^{x-6})$

Using the properties of logarithms, rewrite the equation:

$x \ln(1) = (x - 6) \ln(3)$

Since $\ln(1) = 0$ , the equation simplifies to:

$0 = (x - 6) \ln(3)$

Now, you have two possibilities:

$(x - 6) = 0$ (because $\ln(3) \neq 0$ )
$\ln(3) = 0$ (because $x - 6 \neq 0$ )

Solve these equations:

$x - 6 = 0$ $x = 6$
$\ln(3) = 0$ This is not possible because $\ln(3)$ is not equal to zero.

Therefore, the only solution is $x = 6$ .

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.