Question

Solve exponential equation: $ {100^{7x + 1}} = {1000^{3x - 2}} $

Answer 1

Hint: The problem deals with comparing the powers or indices of two numbers using the basic laws of exponents. Since the two numbers whose powers are to be equated are not the same, we have to first make the bases of exponents the same before comparing the exponents. For making the bases same, we use basic exponent rules.

Complete step-by-step answer:
Making use of laws of exponents to make the bases the same because exponents can be compared only when the bases are equal. We know that $ 100 = {10^2} $ and $ 1000 = {10^3} $ .
So, $ {100^{7x + 1}} = {1000^{3x - 2}} $
 \[ = \] $ {\left[ {{{\left( {10} \right)}^2}} \right] ^{7x + 1}} = {\left[ {{{\left( {10} \right)}^3}} \right] ^{3x - 2}} $ \[\]
Now, using law of exponent $ {\left( {{a^n}} \right)^m} = {\left( a \right)^{nm}} $ , we get
 \[ = \] \[{10^{2(7x + 1)}} = {10^{3(3x - 2)}}\]
 \[ = \] \[{10^{14x + 2}} = {10^{9x - 6}}\]
Now we have the same bases on both sides, so now we can equate exponents or powers of both sides of the equation.
Comparing the exponents,
 \[ = \] $ 14x + 2 = 9x - 6 $
Using transposition rule and shifting $ 9x $ to left side of equation and $ 2 $ to right side of the equation reversing their signs following the sign reversal rule,
 \[ = \] $ 14x - 9x = - 6 - 2 $
On solving further, we get,
 \[ = \] $ 5x = - 8 $
Using transposition rule of algebra and shifting $ 5 $ to right side of the equation,
 \[ = \] $ x = - \dfrac{8}{5} $
So, we get $ x = - \dfrac{8}{5} $ on solving the given exponential equation by equating exponents after making the bases same using laws of exponents.
So, the correct answer is “ $ x = - \dfrac{8}{5} $ ”.

Note: We can also solve the given exponential equation by use of logarithms by taking \[\log \] to the base \[10\] on both sides. Then, solving using simple algebraic rules like transposition, we get the same answer as by the former method. So, exponential equations like the one given in the question can be solved by various methods.