How to solve this? log

Question

How to solve this? log

${(5^{x + 1})}^{2} = 0.2 \sqrt{5^{x}}$ $(5^(x+1))^2 = 0.2sqrt(5^x)$

1 Answer

Impact of this question

1465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-18T15:05:26

$x = - 2$ $x=-2$

Explanation:

Use the rule ${(a^{b})}^{c} = a^{b c}$ $(a^b)^c=a^(bc)$ to simplify the left hand side of the equation. Also, write $\sqrt{5^{x}}$ $sqrt(5^x)$ as ${(5^{x})}^{\frac{1}{2}}$ $(5^x)^(1/2)$ and simplify it using the same rule, to give:

$5^{2 x + 2} = 0.2 (5^{\frac{1}{2} x})$ $5^(2x+2)=0.2(5^(1/2x))$

Now, note that $0.2 = \frac{1}{5} = 5^{- 1}$ $0.2=1/5=5^-1$ .

$5^{2 x + 2} = 5^{- 1} (5^{\frac{1}{2} x})$ $5^(2x+2)=5^-1(5^(1/2x))$

Simplify the right hand side using the rule $a^{b} (a^{b}) = a^{b + c}$ $a^b(a^b)=a^(b+c)$ .

$5^{2 x + 2} = 5^{\frac{1}{2} x - 1}$ $5^(2x+2)=5^(1/2x-1)$

Now, since we have two exponential functions with the same base being equal, we also know that their exponents will be equal.

$2 x + 2 = \frac{1}{2} x - 1$ $2x+2=1/2x-1$

$\frac{3}{2} x = - 3$ $3/2x=-3$

$x = - 3 (\frac{2}{3}) = - 2$ $x=-3(2/3)=-2$

Science

Math

Humanities

... and beyond

How to solve this? log

${(5^{x + 1})}^{2} = 0.2 \sqrt{5^{x}}$ $(5^(x+1))^2 = 0.2sqrt(5^x)$

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How to solve this? *log*

(5x+1)2=0.2√5x(5^(x+1))^2 = 0.2sqrt(5^x)

1 Answer

Explanation:

Related questions

Impact of this question

How to solve this? log

${(5^{x + 1})}^{2} = 0.2 \sqrt{5^{x}}$ $(5^(x+1))^2 = 0.2sqrt(5^x)$