An algebra problem by Mazen Muhammad

Algebra Level 1

${ 3 }^{ x }+{ 3 }^{ \sqrt { x } }=90$

The answer is 4.

2 solutions

Alex Delhumeau
May 30, 2015

$\text{Say }3^{\sqrt{x}}=y$

$\Rightarrow y^2+y-90=0 \Rightarrow y = -10 \text{ or } 9$ .

$-10 \text { is an extraneous root, so ignore it.}$

$3^{\sqrt{x}}=9 \Rightarrow \sqrt{x} = 2 \Rightarrow x = \boxed{4}$ .

Mohd Naim Mohd Amin
Jan 10, 2015

Hey yo...

Let a = 3^x,

a + a^0.5= 90

a^0.5 = 90 - a

Squaring both sides,

a = 8100 - 180a + a^2

a^2 - 181a + 8100 = 0

Factorise the equation,

(a - 81) (a-100) = 0

a = 81 , a = 100

ignore a = 100 , does not fit the original equation,

take a = 81,

3^ x = 81,

3^x = 3^4,

x = 4 , substitute back to check on the originality,

3^4 + 3 ^ 2 = 81 + 9 = 90...

Thanks....

Unfortunately, your substitution does not quite work. If $a = 3^x$ , then $a^{0.5} = 3^{x/2}$ , not $3^{\sqrt{x}}$ .

Jon Haussmann - 6 years, 5 months ago

a^x^y= a^xy

Jon Haussmann

Moklesur Rahaman - 6 years, 5 months ago

It is true that $a^{xy} = (a^x)^y$ . However, it is NOT true in general that $a^{xy} = a^{x^y}$ . (Incidentally, this is why you should use parentheses when you write "a^x^y= a^xy", to make it clear what you mean.)

For example, if $x = 16$ and $a = 3^x = 3^{16}$ , then $\sqrt{a} = (3^{16})^{1/2} = 3^8$ . However, $3^{\sqrt{x}} = 3^{\sqrt{16}} = 3^4$ , so $\sqrt{3^x} \neq 3^{\sqrt{x}}$ .

Jon Haussmann - 6 years, 5 months ago

@Jon Haussmann – you're correct, sir, i will correct it now, thanks for this guide....

MOHD NAIM MOHD AMIN - 6 years, 5 months ago

An algebra problem by Mazen Muhammad

The answer is 4.

2 solutions

Jon Haussmann

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