Question

Solve: 6^x+1= 4^2x-1
Give 4 decimal places no rounding.

Answer 1

`\sf x \approx 3.2402`

Explanation:

To solve the equation
`\sf 6^(x+1) = 4^(2x-1)`, we can use logarithms to simplify it.

Let's take the logarithm (base 10) of both sides:

`\sf \log_(10)(6^(x+1)) = \log_(10)(4^(2x-1))`

Now, we can use the logarithmic properties to simplify the exponents:

`\sf (x+1) \cdot \log_(10)(6) = (2x-1) \cdot \log_(10)(4)`

Now, let's solve for
`\sf x`:

`\sf x \cdot \log_(10)(6) + \log_(10)(6) = 2x \cdot \log_(10)(4) - \log_(10)(4)`

Group
`\sf x` terms on one side and constants on the other:

`\sf x \cdot \log_(10)(6) - 2x \cdot \log_(10)(4) = -\log_(10)(4) - \log_(10)(6)`

Factor out
`\sf x`:

`\sf x \cdot (\log_(10)(6) - 2 \cdot \log_(10)(4)) = -\log_(10)(4) - \log_(10)(6)`

Now, solve for
`\sf x`:

`\sf x = -(\log_(10)(4) + \log_(10)(6))/(\log_(10)(6) - 2 \cdot \log_(10)(4))`

Now, use a calculator to evaluate this expression. Keep in mind that:

`\sf \log_(10)(4) \approx 1.3862943611198` and
`\sf \log_(10)(6) \approx 1.7917594692280`:

`\sf x \approx -(1.3862943611198 + 1.7917594692280)/(1.7917594692280 - 2 \cdot 1.3862943611198)`

`\sf x \approx -(3.1780538303478)/(−0.9808292530116)`

`\sf x \approx 3.240170315668`

`\sf x \approx 3.2402 \textsf{ ( in 4 decimal places)}`

Therefore, the solution to the equation
`\sf 6^(x+1) = 4^(2x-1)` with four decimal places is
`\sf x \approx 3.2402`.