# Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 3)

3.

Given that 10

^{0.48}=*x*, 10^{0.70}=*y*and*x*^{z}=*y*^{2}, then the value of*z*is close to:Answer: Option

Explanation:

*x*^{z} = *y*^{2} 10^{(0.48z)} = 10^{(2 x 0.70)} = 10^{1.40}

0.48*z* = 1.40

z = |
140 | = | 35 | = 2.9 (approx.) |

48 | 12 |

Discussion:

19 comments Page 1 of 2.
Dheeraj verma said:
4 months ago

@All.

Please, can anyone explain in more briefly how we have got 10 (0. 48z) = 10 (2 x 0. 70) = 101. 40?

Please, can anyone explain in more briefly how we have got 10 (0. 48z) = 10 (2 x 0. 70) = 101. 40?

Zaid Shaikh said:
1 year ago

It's like half (0.48) z is 1.40 then full z (0.96) 2.80 therefore 1 Z is 2.9.

(4)

Monisha Mohanan said:
3 years ago

X^ z = y ^ 2

[10^ .048]^z = [10 ^0.70]^2

Taking log on both sides,

Log [10^ .048]^z = log [10 ^0.70]^2

Log a^b^c = c log a^ b

And similarly c log a^b = c*b log a

Z log10^ .048 = 2 log 10 ^0.70

Z * .48 log 10 = 2 * 0.70 log 10

.48 z = 1.4 here log 10 = 1.

Z = 1.4 /.48.

For easy calculation multiplying it by 100,

Z = 140/48 = 2.9.

[10^ .048]^z = [10 ^0.70]^2

Taking log on both sides,

Log [10^ .048]^z = log [10 ^0.70]^2

Log a^b^c = c log a^ b

And similarly c log a^b = c*b log a

Z log10^ .048 = 2 log 10 ^0.70

Z * .48 log 10 = 2 * 0.70 log 10

.48 z = 1.4 here log 10 = 1.

Z = 1.4 /.48.

For easy calculation multiplying it by 100,

Z = 140/48 = 2.9.

Naman said:
4 years ago

I cannot understand it.

(1)

Jesophat said:
5 years ago

48/100 = 0.48z =1.4z.

Akhila said:
6 years ago

@Jaichu.

For both nums decimal at the same place. So, a decimal can be removed.

For both nums decimal at the same place. So, a decimal can be removed.

Aniket Gaikwad said:
7 years ago

Thanks @Tulasi.

Jaichu said:
7 years ago

Please explain me.

0.48z = 1.40.

z =140/48.

How could they multiply it?

0.48z = 1.40.

z =140/48.

How could they multiply it?

Thiyaguaxrev4fvtr said:
8 years ago

How to get the answer? Please explain me.

Kumar arpit said:
9 years ago

That seems correct to me. I think this is the correct way.

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