If $5^{l o g_{a} x} + 5 x^{l o g_{a} 5} = 3$ then value of $x$ is

5^{- l o g_{2} a}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{l o g_{5} a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{l o g_{2} a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5^{l o g_{2} a}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $5^{- l o g_{2} a}$
$5^{{log}_{a} x} + 5 x^{{log}_{a} 5} = 3$
$5^{{log}_{a} x} + 5. 5^{{log}_{a} x} = 3$
$l e t {log}_{a} x = z$
$5^{z} + 5. 5^{z} = 3$
$5^{z} = \frac{1}{2}$
$z = \frac{- log 2}{log 5} = - {log}_{5} 2$
${log}_{a} x = {log}_{5} (\frac{1}{2}) = {log}_{5} (\frac{1}{2})$
$x = 5^{- {log}_{2} a}$

Suggest Corrections

Similar questions

x

5^{{log}_{a} x} + 5 x^{{log}_{a} 5} = 3 (a > 0)

5^{{log}_{a} x} + 5 x^{{log}_{a} 5} = 3 (a > 0, a \neq 1)

5^{{log}_{a} x} + 5 x^{{log}_{a} 5} = 3 (a > 0, a \neq 1)

{log}_{2} (a + b) + {log}_{2} (c + d) \geq 4

a + b + c + d

l o g (2^{a} \times 3^{b} \times 5^{c})

l o g (2^{2} \times 3^{3} \times 5), l o g (2^{6} \times 3 \times 5^{7}), a n d l o g (2 \times 3^{2} \times 5^{4}),

Join BYJU'S Learning Program