How do you solve the exponential equation $100^(7x+1)=1000^(3x-2)$ ?

Question

4447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-13T22:18:50

$x = -8/5$

In an exponential equation: $"base"^"exponent"$

Make both $100$ and $1000$ be a power of $10$ so they both have the same $"base"$

$100 = 10^2$ and $1000 = 10^3$

Use the exponent rule power of a power $(x^m)^n = x^(m*n)$

$(10^2)^(7x+1) = (10^3)^(3x-2)$

Simplify the exponents:

$10^(14x+2) = 10^(9x-6)$

When the bases are equivalent, the exponents are equivalent:
$14x+2 = 9x-6$

Solve for $x$ by adding/subtracting like-terms:
$5x = -8$

So $x = -8/5$

Answer 2 · 2017-03-13T22:20:45

I tried this:

Let us take the log in base $10$ of both sides and use properties of logs:

$log_(10)100^(7x+1)=log _(10)1000^(3x-2)$

$(7x+1)log_(10)100=(3x-2)log_(10)1000$

$(7x+1)*2=(3x-2)*3$

$14x+2=9x-6$

$5x=-8$

$x=-8/5$

How do you solve the exponential equation 100^(7x+1)=1000^(3x-2)100^(7x+1)=1000^(3x-2)?