How do you solve $10^{x - 1} = 100^{2 x - 3}$ $10^(x-1)=100^(2x-3)$ ?

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Answer 1 · 2018-06-18T21:26:25

$x = \frac{5}{3}$ $x=5/3$

Let's rewrite the right side in terms of base- $10$ $10$ . This gives us

$10^{x - 1} = 10^{(2) \cdot (2 x - 3)}$ $10^(x-1)=color(blue)(10^((2)*(2x-3))$

Notice, $10^{2} = 100$ $10^2=100$ , so we didn't change the value of the equation.

$\Rightarrow 10^{x - 1} = 10^{4 x - 6}$ $=>10^(x-1)=10^(4x-6)$

Since we have the same bases, the exponents are equivalent. We can now set up the following equation:

$x - 1 = 4 x - 6$ $x-1=4x-6$

Subtracting $4 x$ $4x$ from both sides gives us

$- 3 x - 1 = - 6$ $-3x-1=-6$

Adding $1$ $1$ to both sides, we get

$- 3 x = - 5$ $-3x=-5$

Lastly, dividing both sides by $- 3$ $-3$ , we get

$x = \frac{5}{3}$ $x=5/3$

Hope this helps!

How do you solve 10x−1=1002x−310^(x-1)=100^(2x-3)?