Question

Solve the exponential equation1/16 = 64^4x-3

Answer 1

Hello there. To solve this question, we'll have to remember some properties about exponential equations.

Given the equation:

`(1)/(16)=64^(4x-3)`

First, we need to convert all the bases into the same. We can choose either 2 or 4 for the base, since 16 and 64 are both powers of them.

Choosing 4 for example, and knowing 16 = 4² and 64 = 4³, we have:

`(1)/(16)=(16)^(-1)=(4^2)^{-1^{}}=4^(-2)`

And

`64^(4x-3)=(4^3)^(4x-3)=4^(3(4x-3))=4^(12x-9)`

Thus the equation turns into:

`4^(-2)=4^(12x-9)`

Since the bases are the same, we can equal the exponents:

`-2=12x-9`

Add 9 on both sides of the equation

`12x=7`

Divide both sides by a factor of 12

`x=(7)/(12)`

== Other way of solving

Same equation, this time we don't need to find equal bases.

Take the logarithm of both sides, any base you want, as long it is positive. Say, the logarithm of base e, or natural log:

`\ln \mleft((1)/(16)\mright)=\ln (64^(4x-3))_{}`

By logarithm rules, we know:

`\log \mleft((a)/(b)\mright)=\log (a)-\log (b)`

and

`\log (a^b)=b\cdot\log (a)`

Knowing 16 = 2^4 and 64 = 2^6, we make:

`\begin{gathered} \ln (1)-\ln (16)=(4x-3)\cdot\ln (64) \\ -4\cdot\ln (2)=(4x-3)\cdot6\cdot\ln (2) \end{gathered}`

Here we used the fact that ln(1) = 0, since 1 = a^0 for any power a not equal to zero.

We can cancel the ln(2) factor on both sides of the equation, by dividing both sides by this factor.

`-4=24x-18`

Add 18 on both sides of the equation

`24x=14`

Divide both sides by a factor of 24 and simplify the fraction

`x=(7)/(12)`