Progressive Base and Powers

Algebra Level 1

For what value of x x does 1 0 x × 10 0 2 x = 100 0 5 ? \large 10^{x} \times 100^{2x}=1000^{5} ?

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Jessey Uche-Nwichi by Jessey Uche-Nwichi , 17, Nigeria

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3 solutions

1 0 x × 10 0 2 x = 100 0 5 10^x \times 100^{2x}=1000^5

1 0 x × 1 0 2 ( 2 x ) = 1 0 3 ( 5 ) 10^x \times 10^{2(2x)}=10^{3(5)}

1 0 x × 1 0 4 x = 1 0 15 10^x \times 10^{4x}=10^{15}

1 0 ( x + 4 x ) = 1 0 15 10^{(x+4x)}=10^{15}

1 0 5 x = 1 0 15 10^{5x}=10^{15}

5 x = 15 5x=15

x = x= 3 \boxed{3}

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Moderator note:

Being able to rewrite equations containing exponents so that all bases are the same can be extremely useful in all situations, since once the equation is expressed as

b q = b r b^q = b^r

as long as b b does not equal -1, 0, or 1, the only way the equality is true is when

q = r . q = r .

We can rewrite this expression as log ⁡ ( 1 0 x . 10 0 2 x ) = log ⁡ ( 100 0 5 ) \log(10^x.100^{2x})=\log(1000^5) , which can be simplified to log ⁡ ( 1 0 x . 1 0 4 x ) = 5 log ⁡ ( 1000 ) \log(10^{x}.10^{4x})=5\log(1000) , and that can be further simplified to log ⁡ ( 1 0 5 x ) = 5 log ⁡ ( 1 0 3 ) \log(10^{5x})=5\log(10^3) . This leads to 5 x = 15 5x=15 . Solving this linear equation yields x = 3 x = 3 .

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1 0 x × 10 0 2 x = 100 0 5 log ⁡ ( 1 0 x × 10 0 2 x ) = log ⁡ ( 100 0 5 ) log ⁡ ( 1 0 x ) + log ⁡ ( 10 0 2 x ) = 5 log ⁡ ( 1000 ) x log ⁡ ( 10 ) + 2 x log ⁡ ( 100 ) = 15 x × 1 + 2 x × 2 = 15 x + 4 x = 15 5 x = 15 x = 3 10^{x} \times 100^{2x}=1000^{5} \\ \log(10^{x} \times 100^{2x}) =\log(1000^{5}) \\ \log(10^{x}) + \log(100^{2x}) = 5\log(1000) \\ x\log(10) + 2x\log(100) = 15 \\ x \times 1 + 2x \times 2 =15 \\ x + 4x = 15 \\ 5x = 15 \\ x=3

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