Let y -log a k X - z -log a x- Find the relation between y and z-A- z - y - kB- z-k2 yC- z-y - k2D- z - yk

The correct option is D z = yk y = log_akx We can also write this logarithum y = log_10x/log_10a^k = log_10x/klog_10a y = 1/k log_10x/log_10a = 1/klog_a ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Method of Intervals

Let y =log a ...

Question

Let y = ${log}_{a^{k}} x$ , z = $l o g_{a}$ x. Find the relation between y and z.

z = y/k

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

z = yk

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

z = k²y

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

z = y/k²

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

Open in App

Solution

The correct option is B
z = yk

y = $l o g_{a}$ kx

We can also write this logarithum

y = $\frac{{log}_{10} x}{{log}_{10} a^{k}}$ = $\frac{{log}_{10} x}{k {log}_{10} a}$

y = $\frac{1}{k}$ $\frac{{log}_{10} x}{{log}_{10} a}$ = $\frac{1}{k} {log}_{a} x$

y = $\frac{1}{k}$ . (z) {z = $l o g_{a} x$ }

z = yk

Suggest Corrections

Similar questions

Q. If

y = a^{(\frac{1}{1 - {log}_{a} x})}

and

z = a^{(\frac{1}{1 - {log}_{a} y})}

, then relation between

x

and

z

Given, $Δ A B C \sim Δ P Q R$ .

Find the values of y and z.

Q. Given that

x + y

varies as

z + \frac{1}{z}

, and that

x - y

varies as

z - \frac{1}{z}

, find the relation between

x

and

z

, provided that

z = 2

when

x = 3

and

y = 1

$Δ A B C \sim Δ P Q R$

Find the values of y and z.

Q. If

A = {x, y, z}

then the relation

R = {(x, x), (y, y), (z, z), (z, x), (z, y)}

is ____

Join BYJU'S Learning Program

Let y = logakx, z = logax. Find the relation between y and z.

The correct option is B z = yk y = logakx We can also write this logarithum y = log10xlog10ak = log10xklog10a y = 1k log10xlog10a = 1klogax y = 1k . (z) {z = logax} z = yk

Let y = ${log}_{a^{k}} x$ , z = $l o g_{a}$ x. Find the relation between y and z.